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 $x^2$ - $y^2$ = 71. Use difference of squares to obtain (x + y) * (x - y) = 71. Since 71 is a prime we conclude that (x + y) = 71 and (x - y) = 1. This gives us x = 36 and y = 35. We can verify that this is correct on substituting these values in the original equation.

(b) It is not too hard to notice that the LHS is $n^2$ + $(n+1)^2$ + $(n*(n+1))^2$ and the RHS is $(n*(n+1)+1)^2$ for n = 2, 3, 4 and 5. We will prove that LHS = RHS for all n in integers in (c).

(c) We expand the LHS to obtain $n^2$ + $n^2$ + 2*n + 1 + $($ n^2 $+n)^2$ = $n^4$ + 2* $n^3$ + 3* $n^2$ + 2*n + 1 on expanding and combining like terms. Now it is easy to verify that this is equal to $($ n^2 $+n+1)^2$ by expanding $n^2$ + n + 1 squared. 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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