Do you need $m$ ?

The coefficient of $x^2$ in the expansion is given by:

$\begin{aligned} \sum_{k=2}^{49} \binom k2 + \binom {50}2 m^2 & = (3n+1)\binom {51}3 \\ \sum_{k=2}^{49} \frac {k(k-1)}2 + \frac {50\cdot 49 m^2}2 & = (3n+1)\frac {51 \cdot 50 \cdot 49}{3\cdot 2} \\ \sum_{k=2}^{49} (k^2-k) + 50\cdot 49 m^2 & = 17 \cdot 50 \cdot 49(3n+1) \\ \frac {49\cdot 50(2\cdot 49+1)}6 - 1 - \frac {49\cdot 50}2 + 1 + 50\cdot 49 m^2 & = 17 \cdot 50 \cdot 49(3n+1) \\ \frac {99}6 - \frac 12 + m^2 & = 17(3n+1) \\ 16 + m^2 & = 51n + 17 \\ m^2 & = 51n + 1 \end{aligned}$

$\implies \min(m) = 1$ , when $n=\boxed{0}$ .

Notice, the final coefficient of $x^2$ in the sum is $\Sigma_{n=2}^{49} \binom{n}{2} + \binom{50}{2} \cdot m^2$ . From observation of the pascal's triangle (And I'm sure there is some property that shows this), we see that if $m=1$ , the sum of these is equal to $\binom{51}{3}$ $1\\1, 1\\1, 2, 1\\1, 3, 3, 1\\1, 4, 6, 4, 1\\1, 5, 10, 10, 5, 1\\1, 6, 15, 20, 15, 6, 1\\...$ Notice, going down the $\binom{n}{2}$ line up to $n=5$ , we get $1+3+6+10=20=\binom{6}{3}$ . This can be applied up to any other $n$ , so now, we can solve for $n$ such that $m=1$ . $\binom{51}{3}=(3n+1)\cdot\binom{51}{3}\\3n+1=1\\3n=0\\n=\boxed{0}$

(1 + 3 + 6 + 10 + ....+ 49C2)=(1/6)(n)(n+1)(n+2), where n=48.

19600 + (35 m)^2 = (3n+1) 20825. So, m^2 = 17(3n+1) – 16 .The smallest n=0.