A mod 2 binomial identity











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I would like to show the following identity: for all $n, q geq 0$,




$$sum_k binom{2k}{4k-2n} binom{n+3q}{2k+2q+1} equiv binom{n+3q}{2q-1} pmod{2}.$$




This has been computer-tested for all $n, q leq 100$.



Remarks:




  • My convention is that the binomial coefficient $binom{m}{n}$ is nonzero only if $m geq n geq 0$.

  • If it helps, in terms of generating functions, the sum on the left hand side is $$[x^n y^{2q-1}] , frac{(1+y)^{n+3q}}{x+x^2+y^2}.$$ Of course, the right hand side is equal to $[y^{2q-1}] , (1+y)^{n+3q}$.

  • The identity is not true without modulo $2$.






combinatorics summation modular-arithmetic binomial-coefficients







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    I would like to show the following identity: for all $n, q geq 0$,




    $$sum_k binom{2k}{4k-2n} binom{n+3q}{2k+2q+1} equiv binom{n+3q}{2q-1} pmod{2}.$$




    This has been computer-tested for all $n, q leq 100$.



    Remarks:




    • My convention is that the binomial coefficient $binom{m}{n}$ is nonzero only if $m geq n geq 0$.

    • If it helps, in terms of generating functions, the sum on the left hand side is $$[x^n y^{2q-1}] , frac{(1+y)^{n+3q}}{x+x^2+y^2}.$$ Of course, the right hand side is equal to $[y^{2q-1}] , (1+y)^{n+3q}$.

    • The identity is not true without modulo $2$.






    combinatorics summation modular-arithmetic binomial-coefficients







    share|cite|improve this question
























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      3





      I would like to show the following identity: for all $n, q geq 0$,




      $$sum_k binom{2k}{4k-2n} binom{n+3q}{2k+2q+1} equiv binom{n+3q}{2q-1} pmod{2}.$$




      This has been computer-tested for all $n, q leq 100$.



      Remarks:




      • My convention is that the binomial coefficient $binom{m}{n}$ is nonzero only if $m geq n geq 0$.

      • If it helps, in terms of generating functions, the sum on the left hand side is $$[x^n y^{2q-1}] , frac{(1+y)^{n+3q}}{x+x^2+y^2}.$$ Of course, the right hand side is equal to $[y^{2q-1}] , (1+y)^{n+3q}$.

      • The identity is not true without modulo $2$.






      combinatorics summation modular-arithmetic binomial-coefficients







      share|cite|improve this question













      I would like to show the following identity: for all $n, q geq 0$,




      $$sum_k binom{2k}{4k-2n} binom{n+3q}{2k+2q+1} equiv binom{n+3q}{2q-1} pmod{2}.$$




      This has been computer-tested for all $n, q leq 100$.



      Remarks:




      • My convention is that the binomial coefficient $binom{m}{n}$ is nonzero only if $m geq n geq 0$.

      • If it helps, in terms of generating functions, the sum on the left hand side is $$[x^n y^{2q-1}] , frac{(1+y)^{n+3q}}{x+x^2+y^2}.$$ Of course, the right hand side is equal to $[y^{2q-1}] , (1+y)^{n+3q}$.

      • The identity is not true without modulo $2$.






      combinatorics summation modular-arithmetic binomial-coefficients




      combinatorics summation modular-arithmetic binomial-coefficients






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      asked Nov 17 at 18:31









      JHF

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          Here is a tentative proof:



          We first notice that by Kummer theorem $binom{2n}{2m} equiv binom{n}{m} bmod 2$, so that it suffices to show that $$ S_q(n) equiv T_q(n) pmod 2$$ where
          begin{align*}
          S_q(n):=&sum_k binom{k}{n-k}binom{n+3q}{2k+2q+1}\
          T_q(n):=&binom{n+3q}{2q-1}.
          end{align*}



          From here, the symbol $sim$ means has the same parity as.



          We have
          $binom{a+1}{b+1}sim binom{a}{b+1}+binom{a}{b}$ but also $binom{a+2}{b+2}sim binom{a+2}{b}+binom{a}{b}$.



          We have
          begin{align*} T_q(n+2)&sim binom{n+3q+2}{2q-1} sim binom{n+3q}{2q-1}+binom{n+3q}{2q-3}\
          &sim T_q(n) +binom{n+3+3(q-1)}{2(q-1)-1} \
          &sim T_q(n) +T_{q-1}(n+3) \
          T_q(n+3)&sim T_{q+1}(n+2)+ T_{q+1}(n)
          end{align*}
          On the other hand, we have
          begin{align*}S_q(n+3)& sim sum_k binom{k}{n+3-k}binom{n+3q+3}{2k+2q+1}\
          &sim sum_k binom{k}{n+2-(k-1)}binom{n+2+3q+1}{2k+2q+1}\
          &sim sum_k binom{k-1}{n+2-(k-1)}binom{n+2+3q+1}{2k+2q+1}\
          &+ sum_k binom{k-1}{n+1-(k-1)}binom{n+2+3q+1}{2k+2q+1}\
          &sim sum_k binom{k-1}{n+2-(k-1)}binom{n+2+3q+1}{2(k-1)+2q+3}\
          &+ sum_k binom{k-1}{n+1-(k-1)}binom{n+2+3q+1}{2k+2q+1}\
          &sim sum_k binom{k-1}{n+2-(k-1)}binom{n+2+3(q+1)-2}{2(k-1)+2(q+1)+1}\
          &+ sum_k binom{k-1}{n+1-(k-1)}binom{n+2+3q+1}{2k+2q+1}\
          & sim sum_k binom{k-1}{n+2-(k-1)}binom{n+2+3(q+1)}{2(k-1)+2(q+1)+1}\
          &+sum_kbinom{k-1}{n+2-(k-1)}binom{n+2+3(q+1)-2}{2(k-1)+2(q+1)-1}\
          &+sum_k binom{k-1}{n+1-(k-1)}binom{n+2+3q+1}{2k+2q+1}\
          &sim S_{q+1}(n+2) +sum_kbinom{k}{n+2-k}binom{n+3(q+1)}{2k+2(q+1)-1}\
          &+sum_k binom{k}{n+1-k}binom{n+3(q+1)}{2k+2(q+1)+1}\
          &sim S_{q+1}(n+2) \&+sum_kbinom{k}{n+2-k}binom{n+3(q+1)}{2(k-1)+2(q+1)+1}\
          &+sum_k binom{k}{n+1-k}binom{n+3(q+1)}{2k+2(q+1)+1}\
          &sim S_{q+1}(n+2)+sum_kbinom{k-1}{n+2-k}binom{n+3(q+1)}{2(k-1)+2(q+1)+1}\
          &+sum_kbinom{k-1}{n-(k-1)}binom{n+3(q+1)}{2(k-1)+2(q+1)+1}\
          &+sum_k binom{k}{n+1-k}binom{n+3(q+1)}{2k+2(q+1)+1}\
          &sim S_{q+1}(n+2)+ S_{q+1}(n) \&+sum_kbinom{k-1}{n+2-k}binom{n+3(q+1)}{2(k-1)+2(q+1)+1}+sum_k binom{k}{n+1-k}binom{n+3(q+1)}{2k+2(q+1)+1}\
          &sim S_{q+1}(n+2)+ S_{q+1}(n)+2sum_k binom{k}{n+1-k}binom{n+3(q+1)}{2k+2(q+1)+1}\
          &sim S_{q+1}(n+2)+ S_{q+1}(n) \&+sum_kbinom{k-1}{n+2-k}binom{n+3(q+1)}{2(k-1)+2(q+1)+1}+sum_k binom{k}{n+1-k}binom{n+3(q+1)}{2k+2(q+1)+1}\
          &sim S_{q+1}(n+2)+ S_{q+1}(n)+2sum_k binom{k}{n+1-k}binom{n+3(q+1)}{2k+2(q+1)+1}\
          S_q(n+3)&sim S_{q+1}(n+2)+ S_{q+1}(n)
          end{align*}



          So the parity of both $ S_q(n)$ and $ T_q(n)$ satisfy the same third order linear recurrence on $n$. Then in order to show that $ S_q(n) sim T_q(n)$, it suffices to show that $ S_q(0) sim T_q(0)$, $ S_q(1) sim T_q(1)$ and $ S_q(2) sim T_q(2)$.



          $S_q(0)-T_q(0)$ is clearly an even integer since
          $$S_q(0)-T_q(0)= binom{3q}{q+1}-binom{3q}{q-1}=2 binom{3q+1}{q-1}.$$



          We have $T_q(1)-S_q(1)= binom{3q+1}{q+2}-binom{3q+1}{q-2}$. After some calculation we find $$T_q(1)-S_q(1)= 2frac{5q+7}{3q+3}binom{3q+3}{q-1}$$ which is an even integer since begin{align*}binom{3q+3}{q-1}(5q+7) &equiv binom{3q+3}{q-1}(2q+4)pmod {3q+3}\
          &equiv-binom{3q+3}{q-1}(q-1)pmod {3q+3}\
          &equiv-binom{3q+2}{q-2}(3q+3)pmod {3q+3}\
          &equiv 0pmod {3q+3}end{align*}



          Also, after some calculation, it can be shown that
          begin{align*}
          T_q(2)-S_q(2) &= binom{3q+2}{q+3}-binom{3q+2}{q-1}-binom{3q+2}{q-3}\
          &= 2 binom{3q+5}{q-1} frac{54+301q+258q^2+59q^3}{(3q+5)(3q+4)(3q+3)}.
          end{align*}
          Then, to complete the proof, we need to show that
          $$ binom{3q+5}{q-1} (54+301q+258q^2+59q^3)equiv 0 pmod {(3q+5)(3q+4)(3q+3)}.$$
          We have $$ (54+301q+258q^2+59q^3) equiv (q-1)(66+47q+5q^2)pmod {(3q+5)(3q+4)(3q+3)}.$$



          Then $$ binom{3q+5}{q-1} (54+301q+258q^2+59q^3) equiv (3q+5)binom{3q+4}{q-2}(66+47q+5q^2)pmod {(3q+5)(3q+4)(3q+3)}.$$
          Then it suffices to show that
          $$ binom{3q+4}{q-2}(66+47q+5q^2)equiv 0 pmod {(3q+4)(3q+3)}.$$
          But $3q+4$ and $3q+3$ are coprime, so we need to show that
          $$ binom{3q+4}{q-2}(66+47q+5q^2)equiv 0 pmod {3q+4}$$
          and
          $$ binom{3q+4}{q-2}(66+47q+5q^2)equiv 0 pmod {3q+3}.$$
          That is
          $$ binom{3q+4}{q-2}(14-q^2)equiv 0 pmod {3q+4} tag1$$
          and
          $$ binom{3q+4}{q-2}(24-q-q^2)equiv 0 pmod {3q+3}.tag2$$



          (1) is equivalent to
          begin{align*} frac{3q+4}{q-2}binom{3q+3}{q-3}(14-q^2)&equiv 0 pmod {3q+4} \
          binom{3q+3}{q-3}(14-q^2)&equiv 0 pmod {q-2}\
          10binom{3q+3}{q-3}&equiv 0 pmod {q-2}\
          10frac{q-2}{3q+4}binom{3q+4}{q-2}&equiv 0 pmod {q-2}\
          10binom{3q+4}{q-2}&equiv 0 pmod {3q+4}.\
          end{align*}



          But the last line holds true indeed, because we know from here that
          begin{align*} binom{3q+4}{q-2}&equiv 0 pmod {frac{3q+4}{gcd(3q+4,q-2)}}\
          &equiv 0 pmod {frac{3q+4}{gcd(10,q-2)}}end{align*}



          There remains to show the validity of (2). We have



          $$ binom{3q+4}{q-2}(24-q-q^2)= (3q+3)binom{3q+4}{q-4} frac{3q+4}{(q-2)(q-3)}(24-q-q^2)$$
          so that we need to show that
          $$ binom{3q+2}{q-4}(3q+4)(24-q-q^2)equiv 0pmod {(q-2)(q-3)}. $$
          That is



          $$ binom{3q+2}{q-4}(3q+4)(24-q-q^2)equiv 0pmod {q-2} $$
          and
          $$ binom{3q+2}{q-4}(3q+4)(24-q-q^2)equiv 0pmod {q-3}. $$
          That is
          $$ 180binom{3q+2}{q-4}equiv 0pmod {q-2} $$
          and
          $$ 156 binom{3q+2}{q-4}equiv 0pmod {q-3}. $$
          That is
          $$ 180frac{(q-2)(q-3)}{(3q+4)(q-2)}binom{3q+4}{q-2}equiv 0pmod {q-2} $$
          and
          $$ 156 frac{q-3}{3q+3}binom{3q+3}{q-3}equiv 0pmod {q-3}. $$



          That is
          $$ 180(q-3)binom{3q+4}{q-2}equiv 0pmod {(3q+4)(3q+3)} $$
          and
          $$ 156binom{3q+3}{q-3}equiv 0pmod {3q+3}. $$
          That is
          $$ 180(q-3)binom{3q+4}{q-2}equiv 0pmod {3q+4} tag3$$
          and
          $$ 180(q-3)binom{3q+4}{q-2}equiv 0pmod {3q+3} tag4$$
          and
          $$ 156binom{3q+3}{q-3}equiv 0pmod {3q+3}. tag5$$



          For (5) it holds true because we have $156=13cdot 12$ a multiple of $12$ and from here, we have
          begin{align*}binom{3q+3}{q-3}&equiv 0pmod {frac{3q+3}{gcd(3q+3,q-3)}} \
          &equiv 0pmod {frac{3q+3}{gcd(12,q-3)}} end{align*}
          For (3) it holds true because we have $180=18cdot 10$ a multiple of $10$ and from here, we have
          begin{align*}binom{3q+4}{q-2}&equiv 0pmod {frac{3q+4}{gcd(3q+4,q-2)}} \
          &equiv 0pmod {frac{3q+2}{gcd(10,q-2)}} end{align*}



          We have seen that there exist an integer $K$, such that $binom{3q+3}{q-3}=Kfrac{3q+3}{gcd(12,q-3)}$. Now, with the same argument from here, we have
          begin{align*}binom{3q+3}{q-2}&equiv 0pmod {frac{3q+3}{gcd(3q+3,q-2)}} \
          &equiv 0pmod {frac{3q+3}{gcd(9,q-2)}} end{align*}
          so there exists an integer $L$ such that $binom{3q+3}{q-2}=Lfrac{3q+3}{gcd(9,q-2)}$.
          Then
          begin{align*} 180(q-3)binom{3q+4}{q-2}&=180(q-3)left(binom{3q+3}{q-2}+binom{3q+3}{q-3}right)\
          &= (3q+3)left( frac{180L(q-3)}{gcd(9,q-2)}+frac{180K(q-3)}{gcd(12,q-3)}right). end{align*}
          The second factor on the right hand side is clearly an integer and this establishes the validity of (4). The proof is finished here.






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          • Great! Also, after showing the recurrence relation, the base cases can also be handled by breaking into cases of $q$ even and $q$ odd and using strong induction.
            – JHF
            Nov 21 at 15:54











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          Here is a tentative proof:



          We first notice that by Kummer theorem $binom{2n}{2m} equiv binom{n}{m} bmod 2$, so that it suffices to show that $$ S_q(n) equiv T_q(n) pmod 2$$ where
          begin{align*}
          S_q(n):=&sum_k binom{k}{n-k}binom{n+3q}{2k+2q+1}\
          T_q(n):=&binom{n+3q}{2q-1}.
          end{align*}



          From here, the symbol $sim$ means has the same parity as.



          We have
          $binom{a+1}{b+1}sim binom{a}{b+1}+binom{a}{b}$ but also $binom{a+2}{b+2}sim binom{a+2}{b}+binom{a}{b}$.



          We have
          begin{align*} T_q(n+2)&sim binom{n+3q+2}{2q-1} sim binom{n+3q}{2q-1}+binom{n+3q}{2q-3}\
          &sim T_q(n) +binom{n+3+3(q-1)}{2(q-1)-1} \
          &sim T_q(n) +T_{q-1}(n+3) \
          T_q(n+3)&sim T_{q+1}(n+2)+ T_{q+1}(n)
          end{align*}
          On the other hand, we have
          begin{align*}S_q(n+3)& sim sum_k binom{k}{n+3-k}binom{n+3q+3}{2k+2q+1}\
          &sim sum_k binom{k}{n+2-(k-1)}binom{n+2+3q+1}{2k+2q+1}\
          &sim sum_k binom{k-1}{n+2-(k-1)}binom{n+2+3q+1}{2k+2q+1}\
          &+ sum_k binom{k-1}{n+1-(k-1)}binom{n+2+3q+1}{2k+2q+1}\
          &sim sum_k binom{k-1}{n+2-(k-1)}binom{n+2+3q+1}{2(k-1)+2q+3}\
          &+ sum_k binom{k-1}{n+1-(k-1)}binom{n+2+3q+1}{2k+2q+1}\
          &sim sum_k binom{k-1}{n+2-(k-1)}binom{n+2+3(q+1)-2}{2(k-1)+2(q+1)+1}\
          &+ sum_k binom{k-1}{n+1-(k-1)}binom{n+2+3q+1}{2k+2q+1}\
          & sim sum_k binom{k-1}{n+2-(k-1)}binom{n+2+3(q+1)}{2(k-1)+2(q+1)+1}\
          &+sum_kbinom{k-1}{n+2-(k-1)}binom{n+2+3(q+1)-2}{2(k-1)+2(q+1)-1}\
          &+sum_k binom{k-1}{n+1-(k-1)}binom{n+2+3q+1}{2k+2q+1}\
          &sim S_{q+1}(n+2) +sum_kbinom{k}{n+2-k}binom{n+3(q+1)}{2k+2(q+1)-1}\
          &+sum_k binom{k}{n+1-k}binom{n+3(q+1)}{2k+2(q+1)+1}\
          &sim S_{q+1}(n+2) \&+sum_kbinom{k}{n+2-k}binom{n+3(q+1)}{2(k-1)+2(q+1)+1}\
          &+sum_k binom{k}{n+1-k}binom{n+3(q+1)}{2k+2(q+1)+1}\
          &sim S_{q+1}(n+2)+sum_kbinom{k-1}{n+2-k}binom{n+3(q+1)}{2(k-1)+2(q+1)+1}\
          &+sum_kbinom{k-1}{n-(k-1)}binom{n+3(q+1)}{2(k-1)+2(q+1)+1}\
          &+sum_k binom{k}{n+1-k}binom{n+3(q+1)}{2k+2(q+1)+1}\
          &sim S_{q+1}(n+2)+ S_{q+1}(n) \&+sum_kbinom{k-1}{n+2-k}binom{n+3(q+1)}{2(k-1)+2(q+1)+1}+sum_k binom{k}{n+1-k}binom{n+3(q+1)}{2k+2(q+1)+1}\
          &sim S_{q+1}(n+2)+ S_{q+1}(n)+2sum_k binom{k}{n+1-k}binom{n+3(q+1)}{2k+2(q+1)+1}\
          &sim S_{q+1}(n+2)+ S_{q+1}(n) \&+sum_kbinom{k-1}{n+2-k}binom{n+3(q+1)}{2(k-1)+2(q+1)+1}+sum_k binom{k}{n+1-k}binom{n+3(q+1)}{2k+2(q+1)+1}\
          &sim S_{q+1}(n+2)+ S_{q+1}(n)+2sum_k binom{k}{n+1-k}binom{n+3(q+1)}{2k+2(q+1)+1}\
          S_q(n+3)&sim S_{q+1}(n+2)+ S_{q+1}(n)
          end{align*}



          So the parity of both $ S_q(n)$ and $ T_q(n)$ satisfy the same third order linear recurrence on $n$. Then in order to show that $ S_q(n) sim T_q(n)$, it suffices to show that $ S_q(0) sim T_q(0)$, $ S_q(1) sim T_q(1)$ and $ S_q(2) sim T_q(2)$.



          $S_q(0)-T_q(0)$ is clearly an even integer since
          $$S_q(0)-T_q(0)= binom{3q}{q+1}-binom{3q}{q-1}=2 binom{3q+1}{q-1}.$$



          We have $T_q(1)-S_q(1)= binom{3q+1}{q+2}-binom{3q+1}{q-2}$. After some calculation we find $$T_q(1)-S_q(1)= 2frac{5q+7}{3q+3}binom{3q+3}{q-1}$$ which is an even integer since begin{align*}binom{3q+3}{q-1}(5q+7) &equiv binom{3q+3}{q-1}(2q+4)pmod {3q+3}\
          &equiv-binom{3q+3}{q-1}(q-1)pmod {3q+3}\
          &equiv-binom{3q+2}{q-2}(3q+3)pmod {3q+3}\
          &equiv 0pmod {3q+3}end{align*}



          Also, after some calculation, it can be shown that
          begin{align*}
          T_q(2)-S_q(2) &= binom{3q+2}{q+3}-binom{3q+2}{q-1}-binom{3q+2}{q-3}\
          &= 2 binom{3q+5}{q-1} frac{54+301q+258q^2+59q^3}{(3q+5)(3q+4)(3q+3)}.
          end{align*}
          Then, to complete the proof, we need to show that
          $$ binom{3q+5}{q-1} (54+301q+258q^2+59q^3)equiv 0 pmod {(3q+5)(3q+4)(3q+3)}.$$
          We have $$ (54+301q+258q^2+59q^3) equiv (q-1)(66+47q+5q^2)pmod {(3q+5)(3q+4)(3q+3)}.$$



          Then $$ binom{3q+5}{q-1} (54+301q+258q^2+59q^3) equiv (3q+5)binom{3q+4}{q-2}(66+47q+5q^2)pmod {(3q+5)(3q+4)(3q+3)}.$$
          Then it suffices to show that
          $$ binom{3q+4}{q-2}(66+47q+5q^2)equiv 0 pmod {(3q+4)(3q+3)}.$$
          But $3q+4$ and $3q+3$ are coprime, so we need to show that
          $$ binom{3q+4}{q-2}(66+47q+5q^2)equiv 0 pmod {3q+4}$$
          and
          $$ binom{3q+4}{q-2}(66+47q+5q^2)equiv 0 pmod {3q+3}.$$
          That is
          $$ binom{3q+4}{q-2}(14-q^2)equiv 0 pmod {3q+4} tag1$$
          and
          $$ binom{3q+4}{q-2}(24-q-q^2)equiv 0 pmod {3q+3}.tag2$$



          (1) is equivalent to
          begin{align*} frac{3q+4}{q-2}binom{3q+3}{q-3}(14-q^2)&equiv 0 pmod {3q+4} \
          binom{3q+3}{q-3}(14-q^2)&equiv 0 pmod {q-2}\
          10binom{3q+3}{q-3}&equiv 0 pmod {q-2}\
          10frac{q-2}{3q+4}binom{3q+4}{q-2}&equiv 0 pmod {q-2}\
          10binom{3q+4}{q-2}&equiv 0 pmod {3q+4}.\
          end{align*}



          But the last line holds true indeed, because we know from here that
          begin{align*} binom{3q+4}{q-2}&equiv 0 pmod {frac{3q+4}{gcd(3q+4,q-2)}}\
          &equiv 0 pmod {frac{3q+4}{gcd(10,q-2)}}end{align*}



          There remains to show the validity of (2). We have



          $$ binom{3q+4}{q-2}(24-q-q^2)= (3q+3)binom{3q+4}{q-4} frac{3q+4}{(q-2)(q-3)}(24-q-q^2)$$
          so that we need to show that
          $$ binom{3q+2}{q-4}(3q+4)(24-q-q^2)equiv 0pmod {(q-2)(q-3)}. $$
          That is



          $$ binom{3q+2}{q-4}(3q+4)(24-q-q^2)equiv 0pmod {q-2} $$
          and
          $$ binom{3q+2}{q-4}(3q+4)(24-q-q^2)equiv 0pmod {q-3}. $$
          That is
          $$ 180binom{3q+2}{q-4}equiv 0pmod {q-2} $$
          and
          $$ 156 binom{3q+2}{q-4}equiv 0pmod {q-3}. $$
          That is
          $$ 180frac{(q-2)(q-3)}{(3q+4)(q-2)}binom{3q+4}{q-2}equiv 0pmod {q-2} $$
          and
          $$ 156 frac{q-3}{3q+3}binom{3q+3}{q-3}equiv 0pmod {q-3}. $$



          That is
          $$ 180(q-3)binom{3q+4}{q-2}equiv 0pmod {(3q+4)(3q+3)} $$
          and
          $$ 156binom{3q+3}{q-3}equiv 0pmod {3q+3}. $$
          That is
          $$ 180(q-3)binom{3q+4}{q-2}equiv 0pmod {3q+4} tag3$$
          and
          $$ 180(q-3)binom{3q+4}{q-2}equiv 0pmod {3q+3} tag4$$
          and
          $$ 156binom{3q+3}{q-3}equiv 0pmod {3q+3}. tag5$$



          For (5) it holds true because we have $156=13cdot 12$ a multiple of $12$ and from here, we have
          begin{align*}binom{3q+3}{q-3}&equiv 0pmod {frac{3q+3}{gcd(3q+3,q-3)}} \
          &equiv 0pmod {frac{3q+3}{gcd(12,q-3)}} end{align*}
          For (3) it holds true because we have $180=18cdot 10$ a multiple of $10$ and from here, we have
          begin{align*}binom{3q+4}{q-2}&equiv 0pmod {frac{3q+4}{gcd(3q+4,q-2)}} \
          &equiv 0pmod {frac{3q+2}{gcd(10,q-2)}} end{align*}



          We have seen that there exist an integer $K$, such that $binom{3q+3}{q-3}=Kfrac{3q+3}{gcd(12,q-3)}$. Now, with the same argument from here, we have
          begin{align*}binom{3q+3}{q-2}&equiv 0pmod {frac{3q+3}{gcd(3q+3,q-2)}} \
          &equiv 0pmod {frac{3q+3}{gcd(9,q-2)}} end{align*}
          so there exists an integer $L$ such that $binom{3q+3}{q-2}=Lfrac{3q+3}{gcd(9,q-2)}$.
          Then
          begin{align*} 180(q-3)binom{3q+4}{q-2}&=180(q-3)left(binom{3q+3}{q-2}+binom{3q+3}{q-3}right)\
          &= (3q+3)left( frac{180L(q-3)}{gcd(9,q-2)}+frac{180K(q-3)}{gcd(12,q-3)}right). end{align*}
          The second factor on the right hand side is clearly an integer and this establishes the validity of (4). The proof is finished here.






          share|cite|improve this answer























          • Great! Also, after showing the recurrence relation, the base cases can also be handled by breaking into cases of $q$ even and $q$ odd and using strong induction.
            – JHF
            Nov 21 at 15:54















          up vote
          3
          down vote



          accepted










          Here is a tentative proof:



          We first notice that by Kummer theorem $binom{2n}{2m} equiv binom{n}{m} bmod 2$, so that it suffices to show that $$ S_q(n) equiv T_q(n) pmod 2$$ where
          begin{align*}
          S_q(n):=&sum_k binom{k}{n-k}binom{n+3q}{2k+2q+1}\
          T_q(n):=&binom{n+3q}{2q-1}.
          end{align*}



          From here, the symbol $sim$ means has the same parity as.



          We have
          $binom{a+1}{b+1}sim binom{a}{b+1}+binom{a}{b}$ but also $binom{a+2}{b+2}sim binom{a+2}{b}+binom{a}{b}$.



          We have
          begin{align*} T_q(n+2)&sim binom{n+3q+2}{2q-1} sim binom{n+3q}{2q-1}+binom{n+3q}{2q-3}\
          &sim T_q(n) +binom{n+3+3(q-1)}{2(q-1)-1} \
          &sim T_q(n) +T_{q-1}(n+3) \
          T_q(n+3)&sim T_{q+1}(n+2)+ T_{q+1}(n)
          end{align*}
          On the other hand, we have
          begin{align*}S_q(n+3)& sim sum_k binom{k}{n+3-k}binom{n+3q+3}{2k+2q+1}\
          &sim sum_k binom{k}{n+2-(k-1)}binom{n+2+3q+1}{2k+2q+1}\
          &sim sum_k binom{k-1}{n+2-(k-1)}binom{n+2+3q+1}{2k+2q+1}\
          &+ sum_k binom{k-1}{n+1-(k-1)}binom{n+2+3q+1}{2k+2q+1}\
          &sim sum_k binom{k-1}{n+2-(k-1)}binom{n+2+3q+1}{2(k-1)+2q+3}\
          &+ sum_k binom{k-1}{n+1-(k-1)}binom{n+2+3q+1}{2k+2q+1}\
          &sim sum_k binom{k-1}{n+2-(k-1)}binom{n+2+3(q+1)-2}{2(k-1)+2(q+1)+1}\
          &+ sum_k binom{k-1}{n+1-(k-1)}binom{n+2+3q+1}{2k+2q+1}\
          & sim sum_k binom{k-1}{n+2-(k-1)}binom{n+2+3(q+1)}{2(k-1)+2(q+1)+1}\
          &+sum_kbinom{k-1}{n+2-(k-1)}binom{n+2+3(q+1)-2}{2(k-1)+2(q+1)-1}\
          &+sum_k binom{k-1}{n+1-(k-1)}binom{n+2+3q+1}{2k+2q+1}\
          &sim S_{q+1}(n+2) +sum_kbinom{k}{n+2-k}binom{n+3(q+1)}{2k+2(q+1)-1}\
          &+sum_k binom{k}{n+1-k}binom{n+3(q+1)}{2k+2(q+1)+1}\
          &sim S_{q+1}(n+2) \&+sum_kbinom{k}{n+2-k}binom{n+3(q+1)}{2(k-1)+2(q+1)+1}\
          &+sum_k binom{k}{n+1-k}binom{n+3(q+1)}{2k+2(q+1)+1}\
          &sim S_{q+1}(n+2)+sum_kbinom{k-1}{n+2-k}binom{n+3(q+1)}{2(k-1)+2(q+1)+1}\
          &+sum_kbinom{k-1}{n-(k-1)}binom{n+3(q+1)}{2(k-1)+2(q+1)+1}\
          &+sum_k binom{k}{n+1-k}binom{n+3(q+1)}{2k+2(q+1)+1}\
          &sim S_{q+1}(n+2)+ S_{q+1}(n) \&+sum_kbinom{k-1}{n+2-k}binom{n+3(q+1)}{2(k-1)+2(q+1)+1}+sum_k binom{k}{n+1-k}binom{n+3(q+1)}{2k+2(q+1)+1}\
          &sim S_{q+1}(n+2)+ S_{q+1}(n)+2sum_k binom{k}{n+1-k}binom{n+3(q+1)}{2k+2(q+1)+1}\
          &sim S_{q+1}(n+2)+ S_{q+1}(n) \&+sum_kbinom{k-1}{n+2-k}binom{n+3(q+1)}{2(k-1)+2(q+1)+1}+sum_k binom{k}{n+1-k}binom{n+3(q+1)}{2k+2(q+1)+1}\
          &sim S_{q+1}(n+2)+ S_{q+1}(n)+2sum_k binom{k}{n+1-k}binom{n+3(q+1)}{2k+2(q+1)+1}\
          S_q(n+3)&sim S_{q+1}(n+2)+ S_{q+1}(n)
          end{align*}



          So the parity of both $ S_q(n)$ and $ T_q(n)$ satisfy the same third order linear recurrence on $n$. Then in order to show that $ S_q(n) sim T_q(n)$, it suffices to show that $ S_q(0) sim T_q(0)$, $ S_q(1) sim T_q(1)$ and $ S_q(2) sim T_q(2)$.



          $S_q(0)-T_q(0)$ is clearly an even integer since
          $$S_q(0)-T_q(0)= binom{3q}{q+1}-binom{3q}{q-1}=2 binom{3q+1}{q-1}.$$



          We have $T_q(1)-S_q(1)= binom{3q+1}{q+2}-binom{3q+1}{q-2}$. After some calculation we find $$T_q(1)-S_q(1)= 2frac{5q+7}{3q+3}binom{3q+3}{q-1}$$ which is an even integer since begin{align*}binom{3q+3}{q-1}(5q+7) &equiv binom{3q+3}{q-1}(2q+4)pmod {3q+3}\
          &equiv-binom{3q+3}{q-1}(q-1)pmod {3q+3}\
          &equiv-binom{3q+2}{q-2}(3q+3)pmod {3q+3}\
          &equiv 0pmod {3q+3}end{align*}



          Also, after some calculation, it can be shown that
          begin{align*}
          T_q(2)-S_q(2) &= binom{3q+2}{q+3}-binom{3q+2}{q-1}-binom{3q+2}{q-3}\
          &= 2 binom{3q+5}{q-1} frac{54+301q+258q^2+59q^3}{(3q+5)(3q+4)(3q+3)}.
          end{align*}
          Then, to complete the proof, we need to show that
          $$ binom{3q+5}{q-1} (54+301q+258q^2+59q^3)equiv 0 pmod {(3q+5)(3q+4)(3q+3)}.$$
          We have $$ (54+301q+258q^2+59q^3) equiv (q-1)(66+47q+5q^2)pmod {(3q+5)(3q+4)(3q+3)}.$$



          Then $$ binom{3q+5}{q-1} (54+301q+258q^2+59q^3) equiv (3q+5)binom{3q+4}{q-2}(66+47q+5q^2)pmod {(3q+5)(3q+4)(3q+3)}.$$
          Then it suffices to show that
          $$ binom{3q+4}{q-2}(66+47q+5q^2)equiv 0 pmod {(3q+4)(3q+3)}.$$
          But $3q+4$ and $3q+3$ are coprime, so we need to show that
          $$ binom{3q+4}{q-2}(66+47q+5q^2)equiv 0 pmod {3q+4}$$
          and
          $$ binom{3q+4}{q-2}(66+47q+5q^2)equiv 0 pmod {3q+3}.$$
          That is
          $$ binom{3q+4}{q-2}(14-q^2)equiv 0 pmod {3q+4} tag1$$
          and
          $$ binom{3q+4}{q-2}(24-q-q^2)equiv 0 pmod {3q+3}.tag2$$



          (1) is equivalent to
          begin{align*} frac{3q+4}{q-2}binom{3q+3}{q-3}(14-q^2)&equiv 0 pmod {3q+4} \
          binom{3q+3}{q-3}(14-q^2)&equiv 0 pmod {q-2}\
          10binom{3q+3}{q-3}&equiv 0 pmod {q-2}\
          10frac{q-2}{3q+4}binom{3q+4}{q-2}&equiv 0 pmod {q-2}\
          10binom{3q+4}{q-2}&equiv 0 pmod {3q+4}.\
          end{align*}



          But the last line holds true indeed, because we know from here that
          begin{align*} binom{3q+4}{q-2}&equiv 0 pmod {frac{3q+4}{gcd(3q+4,q-2)}}\
          &equiv 0 pmod {frac{3q+4}{gcd(10,q-2)}}end{align*}



          There remains to show the validity of (2). We have



          $$ binom{3q+4}{q-2}(24-q-q^2)= (3q+3)binom{3q+4}{q-4} frac{3q+4}{(q-2)(q-3)}(24-q-q^2)$$
          so that we need to show that
          $$ binom{3q+2}{q-4}(3q+4)(24-q-q^2)equiv 0pmod {(q-2)(q-3)}. $$
          That is



          $$ binom{3q+2}{q-4}(3q+4)(24-q-q^2)equiv 0pmod {q-2} $$
          and
          $$ binom{3q+2}{q-4}(3q+4)(24-q-q^2)equiv 0pmod {q-3}. $$
          That is
          $$ 180binom{3q+2}{q-4}equiv 0pmod {q-2} $$
          and
          $$ 156 binom{3q+2}{q-4}equiv 0pmod {q-3}. $$
          That is
          $$ 180frac{(q-2)(q-3)}{(3q+4)(q-2)}binom{3q+4}{q-2}equiv 0pmod {q-2} $$
          and
          $$ 156 frac{q-3}{3q+3}binom{3q+3}{q-3}equiv 0pmod {q-3}. $$



          That is
          $$ 180(q-3)binom{3q+4}{q-2}equiv 0pmod {(3q+4)(3q+3)} $$
          and
          $$ 156binom{3q+3}{q-3}equiv 0pmod {3q+3}. $$
          That is
          $$ 180(q-3)binom{3q+4}{q-2}equiv 0pmod {3q+4} tag3$$
          and
          $$ 180(q-3)binom{3q+4}{q-2}equiv 0pmod {3q+3} tag4$$
          and
          $$ 156binom{3q+3}{q-3}equiv 0pmod {3q+3}. tag5$$



          For (5) it holds true because we have $156=13cdot 12$ a multiple of $12$ and from here, we have
          begin{align*}binom{3q+3}{q-3}&equiv 0pmod {frac{3q+3}{gcd(3q+3,q-3)}} \
          &equiv 0pmod {frac{3q+3}{gcd(12,q-3)}} end{align*}
          For (3) it holds true because we have $180=18cdot 10$ a multiple of $10$ and from here, we have
          begin{align*}binom{3q+4}{q-2}&equiv 0pmod {frac{3q+4}{gcd(3q+4,q-2)}} \
          &equiv 0pmod {frac{3q+2}{gcd(10,q-2)}} end{align*}



          We have seen that there exist an integer $K$, such that $binom{3q+3}{q-3}=Kfrac{3q+3}{gcd(12,q-3)}$. Now, with the same argument from here, we have
          begin{align*}binom{3q+3}{q-2}&equiv 0pmod {frac{3q+3}{gcd(3q+3,q-2)}} \
          &equiv 0pmod {frac{3q+3}{gcd(9,q-2)}} end{align*}
          so there exists an integer $L$ such that $binom{3q+3}{q-2}=Lfrac{3q+3}{gcd(9,q-2)}$.
          Then
          begin{align*} 180(q-3)binom{3q+4}{q-2}&=180(q-3)left(binom{3q+3}{q-2}+binom{3q+3}{q-3}right)\
          &= (3q+3)left( frac{180L(q-3)}{gcd(9,q-2)}+frac{180K(q-3)}{gcd(12,q-3)}right). end{align*}
          The second factor on the right hand side is clearly an integer and this establishes the validity of (4). The proof is finished here.






          share|cite|improve this answer























          • Great! Also, after showing the recurrence relation, the base cases can also be handled by breaking into cases of $q$ even and $q$ odd and using strong induction.
            – JHF
            Nov 21 at 15:54













          up vote
          3
          down vote



          accepted







          up vote
          3
          down vote



          accepted






          Here is a tentative proof:



          We first notice that by Kummer theorem $binom{2n}{2m} equiv binom{n}{m} bmod 2$, so that it suffices to show that $$ S_q(n) equiv T_q(n) pmod 2$$ where
          begin{align*}
          S_q(n):=&sum_k binom{k}{n-k}binom{n+3q}{2k+2q+1}\
          T_q(n):=&binom{n+3q}{2q-1}.
          end{align*}



          From here, the symbol $sim$ means has the same parity as.



          We have
          $binom{a+1}{b+1}sim binom{a}{b+1}+binom{a}{b}$ but also $binom{a+2}{b+2}sim binom{a+2}{b}+binom{a}{b}$.



          We have
          begin{align*} T_q(n+2)&sim binom{n+3q+2}{2q-1} sim binom{n+3q}{2q-1}+binom{n+3q}{2q-3}\
          &sim T_q(n) +binom{n+3+3(q-1)}{2(q-1)-1} \
          &sim T_q(n) +T_{q-1}(n+3) \
          T_q(n+3)&sim T_{q+1}(n+2)+ T_{q+1}(n)
          end{align*}
          On the other hand, we have
          begin{align*}S_q(n+3)& sim sum_k binom{k}{n+3-k}binom{n+3q+3}{2k+2q+1}\
          &sim sum_k binom{k}{n+2-(k-1)}binom{n+2+3q+1}{2k+2q+1}\
          &sim sum_k binom{k-1}{n+2-(k-1)}binom{n+2+3q+1}{2k+2q+1}\
          &+ sum_k binom{k-1}{n+1-(k-1)}binom{n+2+3q+1}{2k+2q+1}\
          &sim sum_k binom{k-1}{n+2-(k-1)}binom{n+2+3q+1}{2(k-1)+2q+3}\
          &+ sum_k binom{k-1}{n+1-(k-1)}binom{n+2+3q+1}{2k+2q+1}\
          &sim sum_k binom{k-1}{n+2-(k-1)}binom{n+2+3(q+1)-2}{2(k-1)+2(q+1)+1}\
          &+ sum_k binom{k-1}{n+1-(k-1)}binom{n+2+3q+1}{2k+2q+1}\
          & sim sum_k binom{k-1}{n+2-(k-1)}binom{n+2+3(q+1)}{2(k-1)+2(q+1)+1}\
          &+sum_kbinom{k-1}{n+2-(k-1)}binom{n+2+3(q+1)-2}{2(k-1)+2(q+1)-1}\
          &+sum_k binom{k-1}{n+1-(k-1)}binom{n+2+3q+1}{2k+2q+1}\
          &sim S_{q+1}(n+2) +sum_kbinom{k}{n+2-k}binom{n+3(q+1)}{2k+2(q+1)-1}\
          &+sum_k binom{k}{n+1-k}binom{n+3(q+1)}{2k+2(q+1)+1}\
          &sim S_{q+1}(n+2) \&+sum_kbinom{k}{n+2-k}binom{n+3(q+1)}{2(k-1)+2(q+1)+1}\
          &+sum_k binom{k}{n+1-k}binom{n+3(q+1)}{2k+2(q+1)+1}\
          &sim S_{q+1}(n+2)+sum_kbinom{k-1}{n+2-k}binom{n+3(q+1)}{2(k-1)+2(q+1)+1}\
          &+sum_kbinom{k-1}{n-(k-1)}binom{n+3(q+1)}{2(k-1)+2(q+1)+1}\
          &+sum_k binom{k}{n+1-k}binom{n+3(q+1)}{2k+2(q+1)+1}\
          &sim S_{q+1}(n+2)+ S_{q+1}(n) \&+sum_kbinom{k-1}{n+2-k}binom{n+3(q+1)}{2(k-1)+2(q+1)+1}+sum_k binom{k}{n+1-k}binom{n+3(q+1)}{2k+2(q+1)+1}\
          &sim S_{q+1}(n+2)+ S_{q+1}(n)+2sum_k binom{k}{n+1-k}binom{n+3(q+1)}{2k+2(q+1)+1}\
          &sim S_{q+1}(n+2)+ S_{q+1}(n) \&+sum_kbinom{k-1}{n+2-k}binom{n+3(q+1)}{2(k-1)+2(q+1)+1}+sum_k binom{k}{n+1-k}binom{n+3(q+1)}{2k+2(q+1)+1}\
          &sim S_{q+1}(n+2)+ S_{q+1}(n)+2sum_k binom{k}{n+1-k}binom{n+3(q+1)}{2k+2(q+1)+1}\
          S_q(n+3)&sim S_{q+1}(n+2)+ S_{q+1}(n)
          end{align*}



          So the parity of both $ S_q(n)$ and $ T_q(n)$ satisfy the same third order linear recurrence on $n$. Then in order to show that $ S_q(n) sim T_q(n)$, it suffices to show that $ S_q(0) sim T_q(0)$, $ S_q(1) sim T_q(1)$ and $ S_q(2) sim T_q(2)$.



          $S_q(0)-T_q(0)$ is clearly an even integer since
          $$S_q(0)-T_q(0)= binom{3q}{q+1}-binom{3q}{q-1}=2 binom{3q+1}{q-1}.$$



          We have $T_q(1)-S_q(1)= binom{3q+1}{q+2}-binom{3q+1}{q-2}$. After some calculation we find $$T_q(1)-S_q(1)= 2frac{5q+7}{3q+3}binom{3q+3}{q-1}$$ which is an even integer since begin{align*}binom{3q+3}{q-1}(5q+7) &equiv binom{3q+3}{q-1}(2q+4)pmod {3q+3}\
          &equiv-binom{3q+3}{q-1}(q-1)pmod {3q+3}\
          &equiv-binom{3q+2}{q-2}(3q+3)pmod {3q+3}\
          &equiv 0pmod {3q+3}end{align*}



          Also, after some calculation, it can be shown that
          begin{align*}
          T_q(2)-S_q(2) &= binom{3q+2}{q+3}-binom{3q+2}{q-1}-binom{3q+2}{q-3}\
          &= 2 binom{3q+5}{q-1} frac{54+301q+258q^2+59q^3}{(3q+5)(3q+4)(3q+3)}.
          end{align*}
          Then, to complete the proof, we need to show that
          $$ binom{3q+5}{q-1} (54+301q+258q^2+59q^3)equiv 0 pmod {(3q+5)(3q+4)(3q+3)}.$$
          We have $$ (54+301q+258q^2+59q^3) equiv (q-1)(66+47q+5q^2)pmod {(3q+5)(3q+4)(3q+3)}.$$



          Then $$ binom{3q+5}{q-1} (54+301q+258q^2+59q^3) equiv (3q+5)binom{3q+4}{q-2}(66+47q+5q^2)pmod {(3q+5)(3q+4)(3q+3)}.$$
          Then it suffices to show that
          $$ binom{3q+4}{q-2}(66+47q+5q^2)equiv 0 pmod {(3q+4)(3q+3)}.$$
          But $3q+4$ and $3q+3$ are coprime, so we need to show that
          $$ binom{3q+4}{q-2}(66+47q+5q^2)equiv 0 pmod {3q+4}$$
          and
          $$ binom{3q+4}{q-2}(66+47q+5q^2)equiv 0 pmod {3q+3}.$$
          That is
          $$ binom{3q+4}{q-2}(14-q^2)equiv 0 pmod {3q+4} tag1$$
          and
          $$ binom{3q+4}{q-2}(24-q-q^2)equiv 0 pmod {3q+3}.tag2$$



          (1) is equivalent to
          begin{align*} frac{3q+4}{q-2}binom{3q+3}{q-3}(14-q^2)&equiv 0 pmod {3q+4} \
          binom{3q+3}{q-3}(14-q^2)&equiv 0 pmod {q-2}\
          10binom{3q+3}{q-3}&equiv 0 pmod {q-2}\
          10frac{q-2}{3q+4}binom{3q+4}{q-2}&equiv 0 pmod {q-2}\
          10binom{3q+4}{q-2}&equiv 0 pmod {3q+4}.\
          end{align*}



          But the last line holds true indeed, because we know from here that
          begin{align*} binom{3q+4}{q-2}&equiv 0 pmod {frac{3q+4}{gcd(3q+4,q-2)}}\
          &equiv 0 pmod {frac{3q+4}{gcd(10,q-2)}}end{align*}



          There remains to show the validity of (2). We have



          $$ binom{3q+4}{q-2}(24-q-q^2)= (3q+3)binom{3q+4}{q-4} frac{3q+4}{(q-2)(q-3)}(24-q-q^2)$$
          so that we need to show that
          $$ binom{3q+2}{q-4}(3q+4)(24-q-q^2)equiv 0pmod {(q-2)(q-3)}. $$
          That is



          $$ binom{3q+2}{q-4}(3q+4)(24-q-q^2)equiv 0pmod {q-2} $$
          and
          $$ binom{3q+2}{q-4}(3q+4)(24-q-q^2)equiv 0pmod {q-3}. $$
          That is
          $$ 180binom{3q+2}{q-4}equiv 0pmod {q-2} $$
          and
          $$ 156 binom{3q+2}{q-4}equiv 0pmod {q-3}. $$
          That is
          $$ 180frac{(q-2)(q-3)}{(3q+4)(q-2)}binom{3q+4}{q-2}equiv 0pmod {q-2} $$
          and
          $$ 156 frac{q-3}{3q+3}binom{3q+3}{q-3}equiv 0pmod {q-3}. $$



          That is
          $$ 180(q-3)binom{3q+4}{q-2}equiv 0pmod {(3q+4)(3q+3)} $$
          and
          $$ 156binom{3q+3}{q-3}equiv 0pmod {3q+3}. $$
          That is
          $$ 180(q-3)binom{3q+4}{q-2}equiv 0pmod {3q+4} tag3$$
          and
          $$ 180(q-3)binom{3q+4}{q-2}equiv 0pmod {3q+3} tag4$$
          and
          $$ 156binom{3q+3}{q-3}equiv 0pmod {3q+3}. tag5$$



          For (5) it holds true because we have $156=13cdot 12$ a multiple of $12$ and from here, we have
          begin{align*}binom{3q+3}{q-3}&equiv 0pmod {frac{3q+3}{gcd(3q+3,q-3)}} \
          &equiv 0pmod {frac{3q+3}{gcd(12,q-3)}} end{align*}
          For (3) it holds true because we have $180=18cdot 10$ a multiple of $10$ and from here, we have
          begin{align*}binom{3q+4}{q-2}&equiv 0pmod {frac{3q+4}{gcd(3q+4,q-2)}} \
          &equiv 0pmod {frac{3q+2}{gcd(10,q-2)}} end{align*}



          We have seen that there exist an integer $K$, such that $binom{3q+3}{q-3}=Kfrac{3q+3}{gcd(12,q-3)}$. Now, with the same argument from here, we have
          begin{align*}binom{3q+3}{q-2}&equiv 0pmod {frac{3q+3}{gcd(3q+3,q-2)}} \
          &equiv 0pmod {frac{3q+3}{gcd(9,q-2)}} end{align*}
          so there exists an integer $L$ such that $binom{3q+3}{q-2}=Lfrac{3q+3}{gcd(9,q-2)}$.
          Then
          begin{align*} 180(q-3)binom{3q+4}{q-2}&=180(q-3)left(binom{3q+3}{q-2}+binom{3q+3}{q-3}right)\
          &= (3q+3)left( frac{180L(q-3)}{gcd(9,q-2)}+frac{180K(q-3)}{gcd(12,q-3)}right). end{align*}
          The second factor on the right hand side is clearly an integer and this establishes the validity of (4). The proof is finished here.






          share|cite|improve this answer














          Here is a tentative proof:



          We first notice that by Kummer theorem $binom{2n}{2m} equiv binom{n}{m} bmod 2$, so that it suffices to show that $$ S_q(n) equiv T_q(n) pmod 2$$ where
          begin{align*}
          S_q(n):=&sum_k binom{k}{n-k}binom{n+3q}{2k+2q+1}\
          T_q(n):=&binom{n+3q}{2q-1}.
          end{align*}



          From here, the symbol $sim$ means has the same parity as.



          We have
          $binom{a+1}{b+1}sim binom{a}{b+1}+binom{a}{b}$ but also $binom{a+2}{b+2}sim binom{a+2}{b}+binom{a}{b}$.



          We have
          begin{align*} T_q(n+2)&sim binom{n+3q+2}{2q-1} sim binom{n+3q}{2q-1}+binom{n+3q}{2q-3}\
          &sim T_q(n) +binom{n+3+3(q-1)}{2(q-1)-1} \
          &sim T_q(n) +T_{q-1}(n+3) \
          T_q(n+3)&sim T_{q+1}(n+2)+ T_{q+1}(n)
          end{align*}
          On the other hand, we have
          begin{align*}S_q(n+3)& sim sum_k binom{k}{n+3-k}binom{n+3q+3}{2k+2q+1}\
          &sim sum_k binom{k}{n+2-(k-1)}binom{n+2+3q+1}{2k+2q+1}\
          &sim sum_k binom{k-1}{n+2-(k-1)}binom{n+2+3q+1}{2k+2q+1}\
          &+ sum_k binom{k-1}{n+1-(k-1)}binom{n+2+3q+1}{2k+2q+1}\
          &sim sum_k binom{k-1}{n+2-(k-1)}binom{n+2+3q+1}{2(k-1)+2q+3}\
          &+ sum_k binom{k-1}{n+1-(k-1)}binom{n+2+3q+1}{2k+2q+1}\
          &sim sum_k binom{k-1}{n+2-(k-1)}binom{n+2+3(q+1)-2}{2(k-1)+2(q+1)+1}\
          &+ sum_k binom{k-1}{n+1-(k-1)}binom{n+2+3q+1}{2k+2q+1}\
          & sim sum_k binom{k-1}{n+2-(k-1)}binom{n+2+3(q+1)}{2(k-1)+2(q+1)+1}\
          &+sum_kbinom{k-1}{n+2-(k-1)}binom{n+2+3(q+1)-2}{2(k-1)+2(q+1)-1}\
          &+sum_k binom{k-1}{n+1-(k-1)}binom{n+2+3q+1}{2k+2q+1}\
          &sim S_{q+1}(n+2) +sum_kbinom{k}{n+2-k}binom{n+3(q+1)}{2k+2(q+1)-1}\
          &+sum_k binom{k}{n+1-k}binom{n+3(q+1)}{2k+2(q+1)+1}\
          &sim S_{q+1}(n+2) \&+sum_kbinom{k}{n+2-k}binom{n+3(q+1)}{2(k-1)+2(q+1)+1}\
          &+sum_k binom{k}{n+1-k}binom{n+3(q+1)}{2k+2(q+1)+1}\
          &sim S_{q+1}(n+2)+sum_kbinom{k-1}{n+2-k}binom{n+3(q+1)}{2(k-1)+2(q+1)+1}\
          &+sum_kbinom{k-1}{n-(k-1)}binom{n+3(q+1)}{2(k-1)+2(q+1)+1}\
          &+sum_k binom{k}{n+1-k}binom{n+3(q+1)}{2k+2(q+1)+1}\
          &sim S_{q+1}(n+2)+ S_{q+1}(n) \&+sum_kbinom{k-1}{n+2-k}binom{n+3(q+1)}{2(k-1)+2(q+1)+1}+sum_k binom{k}{n+1-k}binom{n+3(q+1)}{2k+2(q+1)+1}\
          &sim S_{q+1}(n+2)+ S_{q+1}(n)+2sum_k binom{k}{n+1-k}binom{n+3(q+1)}{2k+2(q+1)+1}\
          &sim S_{q+1}(n+2)+ S_{q+1}(n) \&+sum_kbinom{k-1}{n+2-k}binom{n+3(q+1)}{2(k-1)+2(q+1)+1}+sum_k binom{k}{n+1-k}binom{n+3(q+1)}{2k+2(q+1)+1}\
          &sim S_{q+1}(n+2)+ S_{q+1}(n)+2sum_k binom{k}{n+1-k}binom{n+3(q+1)}{2k+2(q+1)+1}\
          S_q(n+3)&sim S_{q+1}(n+2)+ S_{q+1}(n)
          end{align*}



          So the parity of both $ S_q(n)$ and $ T_q(n)$ satisfy the same third order linear recurrence on $n$. Then in order to show that $ S_q(n) sim T_q(n)$, it suffices to show that $ S_q(0) sim T_q(0)$, $ S_q(1) sim T_q(1)$ and $ S_q(2) sim T_q(2)$.



          $S_q(0)-T_q(0)$ is clearly an even integer since
          $$S_q(0)-T_q(0)= binom{3q}{q+1}-binom{3q}{q-1}=2 binom{3q+1}{q-1}.$$



          We have $T_q(1)-S_q(1)= binom{3q+1}{q+2}-binom{3q+1}{q-2}$. After some calculation we find $$T_q(1)-S_q(1)= 2frac{5q+7}{3q+3}binom{3q+3}{q-1}$$ which is an even integer since begin{align*}binom{3q+3}{q-1}(5q+7) &equiv binom{3q+3}{q-1}(2q+4)pmod {3q+3}\
          &equiv-binom{3q+3}{q-1}(q-1)pmod {3q+3}\
          &equiv-binom{3q+2}{q-2}(3q+3)pmod {3q+3}\
          &equiv 0pmod {3q+3}end{align*}



          Also, after some calculation, it can be shown that
          begin{align*}
          T_q(2)-S_q(2) &= binom{3q+2}{q+3}-binom{3q+2}{q-1}-binom{3q+2}{q-3}\
          &= 2 binom{3q+5}{q-1} frac{54+301q+258q^2+59q^3}{(3q+5)(3q+4)(3q+3)}.
          end{align*}
          Then, to complete the proof, we need to show that
          $$ binom{3q+5}{q-1} (54+301q+258q^2+59q^3)equiv 0 pmod {(3q+5)(3q+4)(3q+3)}.$$
          We have $$ (54+301q+258q^2+59q^3) equiv (q-1)(66+47q+5q^2)pmod {(3q+5)(3q+4)(3q+3)}.$$



          Then $$ binom{3q+5}{q-1} (54+301q+258q^2+59q^3) equiv (3q+5)binom{3q+4}{q-2}(66+47q+5q^2)pmod {(3q+5)(3q+4)(3q+3)}.$$
          Then it suffices to show that
          $$ binom{3q+4}{q-2}(66+47q+5q^2)equiv 0 pmod {(3q+4)(3q+3)}.$$
          But $3q+4$ and $3q+3$ are coprime, so we need to show that
          $$ binom{3q+4}{q-2}(66+47q+5q^2)equiv 0 pmod {3q+4}$$
          and
          $$ binom{3q+4}{q-2}(66+47q+5q^2)equiv 0 pmod {3q+3}.$$
          That is
          $$ binom{3q+4}{q-2}(14-q^2)equiv 0 pmod {3q+4} tag1$$
          and
          $$ binom{3q+4}{q-2}(24-q-q^2)equiv 0 pmod {3q+3}.tag2$$



          (1) is equivalent to
          begin{align*} frac{3q+4}{q-2}binom{3q+3}{q-3}(14-q^2)&equiv 0 pmod {3q+4} \
          binom{3q+3}{q-3}(14-q^2)&equiv 0 pmod {q-2}\
          10binom{3q+3}{q-3}&equiv 0 pmod {q-2}\
          10frac{q-2}{3q+4}binom{3q+4}{q-2}&equiv 0 pmod {q-2}\
          10binom{3q+4}{q-2}&equiv 0 pmod {3q+4}.\
          end{align*}



          But the last line holds true indeed, because we know from here that
          begin{align*} binom{3q+4}{q-2}&equiv 0 pmod {frac{3q+4}{gcd(3q+4,q-2)}}\
          &equiv 0 pmod {frac{3q+4}{gcd(10,q-2)}}end{align*}



          There remains to show the validity of (2). We have



          $$ binom{3q+4}{q-2}(24-q-q^2)= (3q+3)binom{3q+4}{q-4} frac{3q+4}{(q-2)(q-3)}(24-q-q^2)$$
          so that we need to show that
          $$ binom{3q+2}{q-4}(3q+4)(24-q-q^2)equiv 0pmod {(q-2)(q-3)}. $$
          That is



          $$ binom{3q+2}{q-4}(3q+4)(24-q-q^2)equiv 0pmod {q-2} $$
          and
          $$ binom{3q+2}{q-4}(3q+4)(24-q-q^2)equiv 0pmod {q-3}. $$
          That is
          $$ 180binom{3q+2}{q-4}equiv 0pmod {q-2} $$
          and
          $$ 156 binom{3q+2}{q-4}equiv 0pmod {q-3}. $$
          That is
          $$ 180frac{(q-2)(q-3)}{(3q+4)(q-2)}binom{3q+4}{q-2}equiv 0pmod {q-2} $$
          and
          $$ 156 frac{q-3}{3q+3}binom{3q+3}{q-3}equiv 0pmod {q-3}. $$



          That is
          $$ 180(q-3)binom{3q+4}{q-2}equiv 0pmod {(3q+4)(3q+3)} $$
          and
          $$ 156binom{3q+3}{q-3}equiv 0pmod {3q+3}. $$
          That is
          $$ 180(q-3)binom{3q+4}{q-2}equiv 0pmod {3q+4} tag3$$
          and
          $$ 180(q-3)binom{3q+4}{q-2}equiv 0pmod {3q+3} tag4$$
          and
          $$ 156binom{3q+3}{q-3}equiv 0pmod {3q+3}. tag5$$



          For (5) it holds true because we have $156=13cdot 12$ a multiple of $12$ and from here, we have
          begin{align*}binom{3q+3}{q-3}&equiv 0pmod {frac{3q+3}{gcd(3q+3,q-3)}} \
          &equiv 0pmod {frac{3q+3}{gcd(12,q-3)}} end{align*}
          For (3) it holds true because we have $180=18cdot 10$ a multiple of $10$ and from here, we have
          begin{align*}binom{3q+4}{q-2}&equiv 0pmod {frac{3q+4}{gcd(3q+4,q-2)}} \
          &equiv 0pmod {frac{3q+2}{gcd(10,q-2)}} end{align*}



          We have seen that there exist an integer $K$, such that $binom{3q+3}{q-3}=Kfrac{3q+3}{gcd(12,q-3)}$. Now, with the same argument from here, we have
          begin{align*}binom{3q+3}{q-2}&equiv 0pmod {frac{3q+3}{gcd(3q+3,q-2)}} \
          &equiv 0pmod {frac{3q+3}{gcd(9,q-2)}} end{align*}
          so there exists an integer $L$ such that $binom{3q+3}{q-2}=Lfrac{3q+3}{gcd(9,q-2)}$.
          Then
          begin{align*} 180(q-3)binom{3q+4}{q-2}&=180(q-3)left(binom{3q+3}{q-2}+binom{3q+3}{q-3}right)\
          &= (3q+3)left( frac{180L(q-3)}{gcd(9,q-2)}+frac{180K(q-3)}{gcd(12,q-3)}right). end{align*}
          The second factor on the right hand side is clearly an integer and this establishes the validity of (4). The proof is finished here.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Nov 18 at 23:25

























          answered Nov 18 at 22:51









          René Gy

          1,081613




          1,081613












          • Great! Also, after showing the recurrence relation, the base cases can also be handled by breaking into cases of $q$ even and $q$ odd and using strong induction.
            – JHF
            Nov 21 at 15:54


















          • Great! Also, after showing the recurrence relation, the base cases can also be handled by breaking into cases of $q$ even and $q$ odd and using strong induction.
            – JHF
            Nov 21 at 15:54
















          Great! Also, after showing the recurrence relation, the base cases can also be handled by breaking into cases of $q$ even and $q$ odd and using strong induction.
          – JHF
          Nov 21 at 15:54




          Great! Also, after showing the recurrence relation, the base cases can also be handled by breaking into cases of $q$ even and $q$ odd and using strong induction.
          – JHF
          Nov 21 at 15:54


















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