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I want to prove that there are infinitely many Pythagorean quadruples by induction, using the following pattern:

$1^2+2^2+2^2=3^2$

So I take $d=3$ and I want to prove by induction that I can form another quadruple with $d_1=d^2$. The hypothesis would be that $a_1,b_1,c_1,d_1$ satisfy $a_1^2+b_1^2+c_1^2=d_1^2$, and inductive step would be to prove that there are $a,b,c$ such that $a^2+b^2+c^2=(a_1^2+b_1^2+c_1^2)^2$. I've been fooling around with different formulas and modulo by 3, 4, 8 without success. I'm looking for full solution or strong hints.

I'm aware that this isn't optimal proof because there are more triples than I can get using this method, but I want this one.

Or maybe my intuition isn't correct and there aren't always quadruples for $d=3^n$ where $n$ is power of two?

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    Does the question intend the quadruples to be _primitive_?2012-02-23
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    Original question doesn't, but making the problem harder is always the right thing to do, so I'd appreciate answer on that.2012-02-23

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