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Prove that for each positive integer $m$, the number $9\cdot2^m$ can be written as a sum of three squares of positive integers.

I think that induction might be used here.!! I could write the first step, i.e true for $m=1$ but couldn't proceed further. This might be solved in a different way.

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    $\begingroup$ Do you mean $9 \cdot 2^m$? $\endgroup$
    – user61527
    Sep 14 '13 at 17:27
  • $\begingroup$ ya this only.!!! $\endgroup$ Sep 14 '13 at 17:29
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We have $$9\cdot 2^1=18=4^2+1^2+1^2$$ and $$9\cdot 2^2=36=4^2+4^2+2^2.$$ Now, if $$9\cdot 2^m=x^2+y^2+z^2,$$ then $$9\cdot 2^{m+2}=4(x^2+y^2+z^2)=(2x)^2+(2y)^2+(2z)^2,$$ and we are done.

(Note the result also holds with $m=0$, as $9=2^2+2^2+1$. And, yes, if needed, the argument can be formalized as an induction that has two base cases, and deals differently with even and odd values of $m$.)

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Let me just point out that the solution obtained by the induction procedure are in some sense unique. Indeed, considering our equation modulo 4 and assuming that $m\ge 2$ we get that $x^2+y^2+z^2=0(\mod 4)$ and thus $x,y,z=0(\mod 2).$ After you cancel $2's$ the problem will be reduced to the small values.

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