It is well known that $\frac{n}{gcd(n,k)}|{n\choose k}$ $\implies$ for all $0 < n < 113^{114}$, $113|{113^{114}\choose n}$ Now, let $(\frac{a}{b})$ denote the Legendre symbol We have $(\frac{107}{113})(\frac{113}{107}) = 1$ and $(\frac{113}{107}) = (\frac{6}{107}) = (\frac{2}{107})(\frac{3}{107})$ and $(\frac{3}{107})(\frac{107}{3}) = (\frac{3}{107})(\frac{2}{3}) = -(\frac{3}{107}) = -1$ $\implies (\frac{3}{107}) = 1$ and $(\frac{2}{107}) = -1^{\frac{107^2-1}{8}} = -1$ $\implies (\frac{113}{107}) = -1$ $\implies (\frac{107}{113}) = -1$ $\implies 107^{56} \equiv 107^\frac{113-1}{2} \equiv -1 \pmod {113}$ Now, if $0 < n < 113^{114}$ then $107^{56}(m^2-1) + 2m + 3 \equiv 1-m^2+2m+3 \equiv 0 \pmod {113}$ $\implies m^2-2m-4 \equiv (m-1)^2 -5 \equiv \pmod {113}$ $\implies (m-1)^2 \equiv 5 \pmod {113}$ $\implies (\frac{5}{113}) = 1$ $\implies (\frac{113}{5}) = (\frac{3}{5}) = 1$ which is a contradiction so there are no solutions if $0< n < 113^{114}$ if $0 = n$ or $n \geq 113^{114}$ then we have $107^{56}(m^2-1) + 2m + 3 = 0$ It is easy to see that there are no non-negative solutions to this So there does not exist such non-negative pairs (m,n) that satisfies the given equation