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Number theory Difficulty 6.6 National olympiad Prove it

Example 4 Proof: There exist infinitely many positive integer triples (a,b,c)(a, b, c) such that a2+b2,a2+c2,b2+c2a^{2}+b^{2}, a^{2}+c^{2}, b^{2}+c^{2} are all perfect squares.

Solution

We prove that we can construct using Pythagorean triples. Take any Pythagorean triple (x,y,z)(x, y, z) (not necessarily primitive). Let
a=x4y2z2,b=y4x2z2,c=4xyz,a=x\left|4 y^{2}-z^{2}\right|, b=y\left|4 x^{2}-z^{2}\right|, c=4 x y z,

then we have
a2+b2=x2(3y2x2)2+y2(3x2y2)2=x6+3x2y4+3x4y2+y6=(x2+y2)3=(z3)2a2+c2=x2(4y2+z2)2b2+c2=y2(4x2+z2)2\begin{aligned} a^{2}+b^{2} & =x^{2}\left(3 y^{2}-x^{2}\right)^{2}+y^{2}\left(3 x^{2}-y^{2}\right)^{2} \\ & =x^{6}+3 x^{2} y^{4}+3 x^{4} y^{2}+y^{6} \\ & =\left(x^{2}+y^{2}\right)^{3}=\left(z^{3}\right)^{2} \\ a^{2}+c^{2} & =x^{2}\left(4 y^{2}+z^{2}\right)^{2} \\ b^{2}+c^{2} & =y^{2}\left(4 x^{2}+z^{2}\right)^{2} \end{aligned}

Since there are infinitely many Pythagorean triples, there are infinitely many such triples (a,b,c)(a, b, c).
In particular, when x=3,y=4,z=5x=3, y=4, z=5, we get a=117,b=44,c=240a=117, b=44, c=240, and
1172+442=1252,1172+2402=2672,442+2402=2442.117^{2}+44^{2}=125^{2}, 117^{2}+240^{2}=267^{2}, 44^{2}+240^{2}=244^{2} .

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Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic and difficulty added by this site.