# Ask Uncle Colin: It’s Hip To Be Square

Dear Uncle Colin,

I’m struggling to make any headway with this: find all integers $n$ such that $5 \times 2^n + 1$ is square. Any ideas?

Lousy Expression Being Equalto Square Gives Undue Exasperation

Hi, LEBESGUE, and thanks for your message!

Every mathematician should have a Bag Of Tricks – things they look out for that occasionally make problem-solving easier. Think of it as a mathematical toolkit. For me, it’s things like Clever Regrouping, Fermat’s Little Theorem, modulo arithmetic and - more often than the others - the difference of two squares.

It’s not immediately obvious that difference of two squares applies here, but let’s call the square number the expression is equal to, $y$, so that $5 \times 2^n + 1= y^2$. Subtracting 1 from both sides gives $5 \times 2^n = y^2 - 1$ - which is the difference of two squares.

So, $5 \times 2^n = (y + 1)(y-1)$. Unless $n$ is zero (which it clearly isn’t), the left-hand-side is even - which makes both of the factors on the right even.

And, in particular, one of them must be a multiple of 4; the other must not be. Looking at the factors of $5\times 2^n$, the only even ones that aren’t multiples of 4 are 2 and 10. We only have two cases to check!

If 2 is one of the factors, the other must be 4 or 0, neither of which can be written in the form $5 \times 2^k$, so that’s no good.

If 10 is one of the factors, the other could be 12 (which is not a power of 2) or 8, which is.

That means $y=9$, making $5 \times 2^n + 1 = 81$. Therefore, $2^n = 16$ and $n=4$.

Hope that helps!

- Uncle Colin ## Colin

Colin is a Weymouth maths tutor, author of several Maths For Dummies books and A-level maths guides. He started Flying Colours Maths in 2008. He lives with an espresso pot and nothing to prove.

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