# Proving that there are infinitely many Pythagorean quadruples by induction using described method

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? –  Aryabhata Feb 23 '12 at 1:41
Original question doesn't, but making the problem harder is always the right thing to do, so I'd appreciate answer on that. –  Lazar Ljubenović Feb 23 '12 at 1:46

We can use the identity

$$n^2 + (n+1)^2 + (n^2 + n)^2 = (n^2 + n+1)^2$$

and generate infinite primitive quadruples. In fact, any given number can be made part of a quadruple!

You can multiply your solution $(1,2,2)$ by $n$ and you get $n^2+(2n)^2+(2n)^2=n^2\cdot(1^2+2^2+2^2)=n^2\cdot 3^2=(3n)^2$.
Oh, so I just take $n_i=3^i$ and I get what I was trying to prove. Now this looks too easy and I've been having so much trouble. Thanks. –  Lazar Ljubenović Feb 23 '12 at 1:39
This does show there are infinitely many quadruples, and it does show there's always one with $d=3^n$. It doesn't use induction (and it doesn't produce primitive triples, if that's what you're after). –  Gerry Myerson Feb 23 '12 at 2:04