Dear Uncle Colin,

I’m told that $5\times 2^x + 1$ (with $x$ a non-negative integer) is a square number - how do I find $x$?

- A Baffling Equation. Logs?

Hi, ABEL, and thanks for your message!

We’re looking for a square number - let’s call it $y^2$ - that’s one more than five times a power of 2. Before I do anything else, I’m going to eliminate $x=0$ as a possible solution: 5+1 is not a square number, so $x>0$.

OK, with that out of the way, we can rewrite our equation as $5 \times 2^x = y^2 - 1$, and factorise the right hand side: $5 \times 2^x = (y+1)(y-1)$.

Because the left-hand side is even (since $x>0$), $y$ must be odd. It must also be either one more or less than a power of two, and one less or more than five times a power of two.

Now, every odd number is sandwiched between two even numbers, exactly one of which is a multiple of 4: so either the “power of two” must be $2^1$ or the “five times a power of two” must be 10.

If one of the factors is 2, the other is either 0 or 4, neither of which gives a valid solution - there isn’t a 5 to be seen!

If one of the factors is 10, the other is either 8 or 12 – and 8 is great, it’s $2^3$.

So, if $y=9$, then we have $\br{ 5\times 2^1 } \br{2^3} = 9^2 - 1$, which we can rewrite as $5\times 2^4 + 1 = 9^2$.

Therefore, the only solution is $x=4$.

Hope that helps!

- Uncle Colin