Simplifying radicals is strangely satisfying.

I’m helping my son with mathematics, and I have to re-learn it, but this is fun. Hence, I’m hoping to start a series of articles.

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Problem statement #

Given \(\begin{align*} A &= \sqrt{4 + \sqrt{11} + \sqrt{23 - 2\sqrt{132}}} \\ B &= \sqrt{\sqrt{97 + 56\sqrt{3}}} \end{align*}\)

Show that: \(\frac{B^2 - 4A}{B - A} \in \mathbb{N}.\)

Solution #

Starting with:

\[B = \sqrt{\sqrt{97 + 56\sqrt{3}}}\]

Often such sums can be rewritten like this:

\[97 + 56\sqrt{3} = (\sqrt{a} + \sqrt{b})^2\]

We observe the formula:

\[(\sqrt{a} + \sqrt{b})^2 = a + b + 2\sqrt{ab}\]

Therefore, we end up with the following system of equations:

\[\begin{cases} a + b = 97 \\ ab = 28^2 \cdot 3 \end{cases}\]

The solutions may or may not jump at you, but there’s a little Vieta’s formula that can help — we can find a and b as being the solutions to this quadratic equation:

\[\begin{align*} (x - a)(x + b) &= 0 \iff \\ x^2 - (a + b)x + ab &= 0 \iff \\ x^2 - 97 x + 2352 &= 0 \end{align*}\]

Solving that:

\[\begin{align*} \Delta &= 97^2 - 4*2352 \\ &= 9409 - 9408 \\ &= 1 \\ x_1 &= a = \frac{97 + \sqrt{1}}{2} = 49 \\ x_2 &= b = \frac{97 - \sqrt{1}}{2} = 48 \end{align*}\]

Threfore:

\[\begin{align*} 97 + 56\sqrt{3} &= (\sqrt{49} + \sqrt{48})^2 \\ &= (7 + 4\sqrt{3})^2 \end{align*}\]

Going back to our original value, we rewrite it:

\[\begin{align*} B &= \sqrt{\sqrt{97 + 56\sqrt{3}}} \\ &= \sqrt{\sqrt{(7 + 4\sqrt{3})^2}} \\ &= \sqrt{7 + 4\sqrt{3}} \end{align*}\]

We still have a square of a sum, but this time we can spot the $(a + b)^2$ formula easily:

\[\begin{align*} B &= \sqrt{7 + 4\sqrt{3}} \\ &= \sqrt{2^2 + 2 * 2\sqrt{3} + \sqrt{3}^2} \\ &= \sqrt{(2 + \sqrt{3})^2} \\ &= 2 + \sqrt{3} \end{align*}\]

Next, for simplifying A, we’ll focus on rewriting this:

\[23 - 2\sqrt{132} = (\sqrt{a} - \sqrt{b})^2\]

Same trick applies, so we’re dealing with this system:

\[\begin{cases} a + b = 23 \\ ab = 132 \end{cases}\]

Which will have solutions generated by this quadratic equation:

\[x^2 - 23x + 132 = 0\]

Solving it:

\[\begin{align*} \Delta &= 23^2 - 4 * 132 = 529 - 528 = 1 \\ x_1 &= a = \frac{23 + 1}{2} = \frac{24}{2} = 12 \\ x_2 &= b = \frac{23 - 1}{2} = \frac{22}{2} = 11 \end{align*}\]

Therefore our sum can be rewritten as (we remember that we have a minus sign though 🙂):

\[\begin{align*} 23 - 2\sqrt{132} &= (\sqrt{12} - \sqrt{11})^2 \\ &= (2\sqrt{3} - \sqrt{11})^2 \end{align*}\]

We can now rewrite our A:

\[\begin{align*} A &= \sqrt{4 + \sqrt{11} + \sqrt{23 - 2\sqrt{132}}} \\ &= \sqrt{4 + \sqrt{11} + \sqrt{(2\sqrt{3} - \sqrt{11})^2}} \\ &= \sqrt{4 + \sqrt{11} + 2\sqrt{3} - \sqrt{11}} \\ &= \sqrt{4 + 2\sqrt{3}} \\ &= \sqrt{1^2 + 2\sqrt{3} + \sqrt{3}^2} \\ &= \sqrt{(1 + \sqrt{3})^2} \\ &= 1 + \sqrt{3} \end{align*}\]

And finally:

\[\begin{align*} \frac{B^2 - 4A}{B - A} &= \frac{(2 + \sqrt{3})^2 - 4(1 + \sqrt{3})}{(2 + \sqrt{3}) - (1 + \sqrt{3})} \\ &= \frac{4 + 4\sqrt{3} + 3 - 4 - 4\sqrt{3}}{1} \\ &= 3 \in \mathbb{N} \end{align*}\]

Published: October 15, 2025 | Written by Alexandru Nedelcu