A nice puzzle from reddit:

Evaluate $\sqrt{\sqrt{2025}-\sqrt{2024}}$

The suggestion is that this is the sort of thing that’ll be in Olympiad papers in a couple of years. Fair enough.

This isn’t quite a real-time solution, but it’s roughly how I thought about it and tackled it. Spoilers are (as you might surmise) below the line.

The first thing I notice is that 2025 is a square number (it’s $45^2$). That means 2024 is $(44)(46)$, although, as it turns out, that’s not immediately helpful to me.

Instead, I followed a problem-solving strategy that was crystallised to me by @colinthemathmo: “make it really big or make it really small”. Could I adapt the problem to one with easier numbers? Let’s consider $x=\sqrt{\sqrt{4}-\sqrt{3}}$, which is more my pace.

I can rewrite that as $x^2 = 2-\sqrt{3}$, and ask “what’s the square root of that?” It’s time for an aside.

### Aside

Let’s suppose $x = a + b\sqrt{3}$, so that $x^2 = \left(a^2 + 3b^2\right) + 2ab\sqrt{3}$.

That leads us to the simultaneous equations $a^2 + 3b^2 = 2$ and $2ab = -1$.

Do you want to know my trick for these? I guess you wouldn’t be reading if you didn’t. It’s “combine them so the loose number is 0.” If I double the second and add it to the first, I get $a^2 + 4ab + 3b^2 = 0$.

That factorises immediately as $(a+b)(a+3b)=0$ – so either $a = -b$ or $a = -3b$.

Case 1, $a=-b$: We’ve got $a^2 + 3(-a)^2 = 2$, so $a^2 = \frac{1}{2}$. That give $a = \pm \sqrt{\frac{1}{2}}$ and $b = \mp \sqrt{\frac{1}{2}}$; this means $x = \left(\sqrt{\frac{3}{2}}-\sqrt{\frac{1}{2}}\right)$, given that we want the positive value.

Case 2, $a=-3b$: Here, $(-3b)^2 + 3b^2 = 2$, so $b^2 = \frac{1}{6}$. We end up with $b = \pm \sqrt{\frac{1}{6}}$ and $a = \mp \sqrt{\frac{3}{2}}$. That leads to the same value of $x$ in the end. This is reassuring; a number can only have two square roots.

OK: we can get a simple enough answer for the case based on 4. A more gung-ho mathematician would hypothesis from here, but I’d like a bit more evidence first. Let’s try $x = \sqrt{\sqrt{100} - \sqrt{99}}$.

We can follow the same approach with a $\sqrt{99}$ instead of a $\sqrt{3}$ and get $a^2 + 99b^2 = 10$ and $2ab = -1$ again.

Skipping the details, which are much the same, we end up with $x = \left(\sqrt{\frac{11}{2}} - \sqrt{\frac{9}{2}}\right)$.

### Now we have enough for a hypothesis

I hypothesise that $\sqrt{n - \sqrt{n^2-1}}$ is $\sqrt{\frac{n+1}{2}}-\sqrt{\frac{n-1}{2}}$, based on two calculations. Let’s prove it:

• $\left(\sqrt{\frac{n+1}{2}}-\sqrt{\frac{n-1}{2}}\right)^2 = \frac{n+1}{2} + \frac{n-1}{2} - 2\left(\sqrt{\frac{n+1}{2}\cdot \frac{n-1}{2}}\right)$
• $\dots = n - \sqrt{(n+1)(n-1)}$
• $\dots = n - \sqrt{n^2-1}$.

Since the second expression is positive for $n \ge 1$, it’s the positive square root of the first, as required.

In our puzzle, $n=45$, so our answer needs to be $\sqrt{23}-\sqrt{22}$.