# 1993 UNCO Math Contest II Problems/Problem 6

## Problem

Observe that \begin{align*} 2^2+3^2+6^2 &= 7^2 \\ 3^2+4^2+12^2 &= 13^2 \\ 4^2+5^2+20^2 &= 21^2 \\ \end{align*}

(a) Find integers $x$ and $y$ so that $5^2+6^2+x^2=y^2.$

(b) Conjecture a general rule that is being illustrated here.

## Solution

(a) We can rewrite the given equation as $y^2-x^2 = 25+36 = 61$. Use difference of squares to obtain $(y + x)(y - x) = 61$. Since $61$ is a prime we conclude that $(y + x) = 61 \text{ and } (y - x) = 1$, giving us $x = 30 \text{ and } y = 31$.

(b) It is not too hard to notice that the LHS above is $n^2 + (n+1)^2 + (n(n+1))^2$ and the RHS above is $(n(n+1)+1)^2$ for $n = 2, 3, 4 \text{ and } 5$. We will prove that the LHS $=$ RHS for all integers (although the proof extends to real numbers) in (c).

(c) We expand the LHS to obtain \begin{align*} n^2 + n^2 + 2n + 1 + n^2(n^2 + 2n + 1) &= n^4 + 2n^3 + 3n^2 + 2n + 1 \\ &= (n^4 + n^3 + n^2) + (n^3 + n^2 + n) + (n^2 + n + 1) \\ &= (n^2 + n + 1)^2 \\ \end{align*} Thus LHS = RHS and we are done. ~AK2006