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.


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 $a\cdot b=5\cdot 11=55$.

  • $\begingroup$ How did the prime factorization help you to split it as sums of squares? Or you are using the factorization just to say 650 is not prime. Is there a systematic way to get those splits, like the OP is asking? $\endgroup$ – Mathemagical Sep 12 '17 at 9:52
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    $\begingroup$ I wrote the factorization just to say that $650$ is not a perfect square. Anyway note that $(a^2+b^2)(c^2+d^2)=(ac+bd)^2+(ad-bc)^2$. $\endgroup$ – Robert Z Sep 12 '17 at 9:56
  • $\begingroup$ Thanks. That's neat. Btw, I meant "650 is not square" (not "prime") in my original comment. $\endgroup$ – Mathemagical Sep 12 '17 at 12:58

For this particular problem, I think the easiest thing is to note that there are $650$ unit squares, which is not a square number. Then subtract $1^2$, $2^2$, etc, from $650$ until the results are square numbers. For $650-5^2$ you get $625$, a square number $25^2$, and for $650-11^2$ you get $529$, a square number $23^2$, so one has $5$ and one has $11$, a product of $55$.

  • $\begingroup$ Please read Mathjax, so other users can properly understand your answer. $\endgroup$ – Mr Pie Sep 15 '17 at 0:41
  • $\begingroup$ Agreed. Sorry, I don't know Mathjax (never heard of it before) - I'll look for a tutorial. $\endgroup$ – Jud McCranie Sep 15 '17 at 3:55

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