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Let $x, y, z$ be integers with $2 < x < y < z$, and let $z$ be even.

If $x^2 + y^2 = z^2 + 4$, what is the least possible value for $x+y+z$ ?

So far, I've realized that $z^2 + 4$ must be even, which means that $x^2 + y^2$ must be even, which means that $x$ and $y$ can either both be even or both be odd.

I'm pretty much stuck at this point. How could you solve this question without resorting to using programming?

Thanks.

amWhy
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user491229
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  • Have you read about Lagrangians and contrained optimisation? – anderstood Oct 18 '17 at 23:38
  • Programming is the natural approach for this problem. It quickly finds the solution $$(x,y,z)=(8, 14, 16)$$ and there are only $5$ other solutions with $z < 40$, all of which have larger sums. – quasi Oct 18 '17 at 23:42
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    @anderstood: I could be mistaken, but I don't see how constrained optimization helps here, since the solutions must be integers. One dumb observation is that we must have $(y-2)(y+2) = (z-x)(z+x)$, so you can enumerate a list of squares minus $4$ and see which of these can be described as a difference of squares. Another small optimization is that $x$ and $y$ must in fact be even, since taking both sides mod $4$ gives $0$ on the RHS, as $z$ is even. – Alex Wertheim Oct 18 '17 at 23:46
  • @AlexWertheim My bad I had missed the integer part, sorry. – anderstood Oct 19 '17 at 02:31
  • https://math.stackexchange.com/questions/74931/integral-solutions-of-x2y21-z2/789972#789972 – individ Oct 19 '17 at 04:24

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