Triangle Squares

We want $\tfrac12n(n+1) = m^2$ , and this equation becomes $(2n+1)^2 - 2(2m)^2 \; = \; 1$ The non-negative integer solutions of Pell's equation $x^2 - 2y^2 = 1$ are obtained by the formula $x_k + y_k\sqrt{2} = (1 + \sqrt{2})^{2k} \; = \; (3 + 2\sqrt{2})^k$ for all $k \ge 0$ . Note that $x_k$ will always be odd, and $y_k$ will always be even. With $x_5 = 3363$ and $y_5 = 2378$ we obtain $n = 1681$ and $m = 1189$ , making the answer $\tfrac12n(n+1) = \boxed{1413721}$

Were looking for a pair of natural numbers $(a,b)$ such that $a^2=\frac{b\left(b+1\right)}{2}$ In order to apply Vieta root-jumping , we'll make $a=b-k$ . Substituting and reorganizing, we have $b^2-4bk-b+2k^2=0$ If we have a solution $(b_1,k_1)$ we can fix $b=b_1$ , which gives us $2k^2-\left(4b_1\right)k+\left(b_1^2-b_1\right)=0$ And from Vieta's formula applied to the constant term, we know that if $k_1$ is an integer solution, then there exists another integer solution $k_2$ such that $k_2=\frac{b_1^2-b_1}{2k_1}$ In an analogous procedure, fixing $k=k_2$ , and reorganizing the equation so it becomes a polinomyal in b, we will find soluutions $b_1$ (which we already found earlier) and $b_2$ such that $b_2=\frac{2k_2^2}{b_1}$ Now, the reason we first fixed $b=b_1$ instead of $k=k_1$ is we want to get greater and greater solutions, by fixing $b=b_1$ first, we would end up at the solution $b=0,k=0$ , you can try it for yourself and check it.

For the same reason, the "intermediate" solution $(b_1,k_2)$ doesn't work for us, because $k_2-b_1=b_1-k_1$ , and remember, $a=b-k$ .

Both previous statements seem a bit out of the blue, but they become obvious if you graph $2y^2-\left(4x\right)y+\left(x^2-x\right)=0$ along with $y=x$ and look at where these solutions are, remember: vieta jumping finds solutions by jumping in ${ 90 }^{ o }$ angles between the hyperbola's "arms".

Summarizing what we've done: if we have a solution $(b_1,k_1)$ , we can find another solution $(b_2,k_2)$ , where $b_2$ and $k_2$ are given by the expressions discussed earlier. Now, we can put $b_3=8$ and $k_3=2$ and jump between the solutions until we get to $(b_6,k_6)$ . I've done this and organized the values found in this table:

Finally, the number we want is $(b_6-k_6)^2$ which is $1189^2$ which is