<s>[INST] Using Square Numbers... To Make Squares I came across a question in an exam today (don't worry, I have already completed the test), and was wondering whether there was a better solution to mine.
The question, as I remember, is as follows: 

The teacher gives 12 students squares of side lengths 1, 2, 3, 4 ... 11, 12. All students receive a distinct square. Then, he asks the students to cut up the squares into unit squares (of side length 1). He challenges the students to arrange their squares adjacently to create a larger square, with no gaps. Of course, they find that it is impossible. 
Alice has a square of side length $a$. She exclaims that if she doesn't use any of her unit squares, then the class is able to create a square.
Similarly, Bill has a square of side length $b$, and says the same thing.
Given neither Alice nor Bill is lying, what is the value of $ab$?

That was a mouthful of a question, and I have attempted it, with very messy results. This is how I started:
\begin{align}
1^2 + 2^2 + 3^2 ... 11^2 + 12^2 -a^2 & = x^2 \\
1^2 + 2^2 + 3^2 ... 11^2 + 12^2 -b^2 & = y^2 \\
\end{align}
Then, by subtracting the latter equation from the first,
\begin{align}
b^2 - a^2 & = x^2 - y^2 \\
(b+a)(b-a) & = (x+y)(x-y) \\
\end{align}
From there, my brain switched off and I just turned to trial & error (which is one of my favourite problem solving techniques).
We know: $1^2 + 2^2 + 3^2 ... 11^2 + 12^2 = 650$, therefore the square created by the class must be $650 - 12^2 < x^2 < 650$.
I got the answers 5 and 11 which multiply for 55, which I believe is correct, but can someone please show me a 'proper' way of completing this question.
 [/INST] Hint. Note that $650=2\cdot 5^2\cdot 13$ (it is not a perfect square) can be written as a sum of $2$ squares in three ways:
$$650=5^2+25^2=11^2+23^2=17^2+19^2.$$
(actually, in your case, it suffices to check for representations $x^2+y^2$ such that $1\leq x\leq 12\leq y$). 
Since $5$ and $11$ are $\leq 12$, it follows that $a\cdot b=5\cdot 11=55$. 
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