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.

(c) Prove your conjecture.

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

1993 UNCO Math Contest II (Problems • Answer Key • Resources)
Preceded by Problem 5	Followed by Problem 7
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1993 UNCO Math Contest II Problems/Problem 6

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