2024 SSMO Relay Round 3 Problems/Problem 2

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Problem

Let $T = TNYWR.$ Find the greatest odd integer $n$ for which $n^2+(T-1)n$ is a perfect square.

Solution

We have $n^2+99n = s^2,$ for some integer $s.$ Factoring, we have \[\left(\left(n+\frac{99}{2}\right)+s\right)\left(\left(n+\frac{99}{2}\right)-s\right)=\frac{99^2}{4}\implies\]\[(2n+99-2s)(2n+99+2s) = 99^2.\] To maximize $n,$ we let $2n+99-2s = 1$ and $2n+99+2s = 99^2 = 9801.$ This gives $n = \frac{\frac{9801+1}{2}-99}{2} = \boxed{2401}.$

~SMO_Team