<s>[INST] Induction Squares Proof Prove that for every positive integer $n$ there exist positive integers $$a_{11}, a_{21}, a_{22}, a_{31}, a_{32}, a_{33}, \dots ,a_{n1}, a_{n2},\dots,a_{nn}$$
such that
$$
a_{11}^2 = a_{21}^2 + a_{22}^2 = a_{31}^2 + a_{32}^2 + a_{33}^2 = a_{n1}^2 + a_{n2}^2 + \cdots + a_{nn}^2.
$$
We're doing a chapter on proofs by induction so I'm pretty sure that would be the way to go. My general thought is to somehow prove that a square exists that can be the sum of any number of squares but I'm not too sure. Thank you for the help!
 [/INST] We proof by induction. You have already found the initial case.
$$5^2 = 3^2 + 4^2$$
We now use the formula
$$\left(\frac{x^2+1}2\right)^2=x^2+\left(\frac{x^2-1}2\right)^2$$ to find a sequence
$$z^2_n=x^2_n+y^2_n$$
such that $x_n=z_{n-1}$.
$$\begin{array}{r,r,r}
5^2&=&4^2&+&3^2\\
13^2&=&12^2&+&5^2\\
85^2&=&84^2&+&13^2\\
3613^2&=&3612^2&+&85^2\\
6 526 885^2&=&6 526 884^2&+&3613^2\\
&\cdots&&&&&
\end{array}$$
From this we get
$$\begin{eqnarray}
6 526 885^2\\
=6 526 884^2&+&3613^2\\
=6 526 884^2&+&3612^2+85^2\\
=6 526 884^2&+&3612^2+84^2+13^2\\
=6 526 884^2&+&3612^2+84^2+12^2+5^2\\
=6 526 884^2&+&3612^2+84^2+12^2+4^2+3^2
\end{eqnarray}
$$
So we have found a number that can be represented as the  sum of 1, 2, 3, 4, 5 or 6 squares. I think it is clear to continue this process and how to proof it by induction.
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