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!

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.
