# How to make a perfect square from a number given in a surd form $a+b\sqrt{c}$?

Is there a way of checking that a number can be written as a perfect square and hence finding it if the number is given in the surd form?

For example, if I expand and simplify $$(1+\sqrt{2})^{2}=3+2\sqrt{2}$$.

Is there a way of finding that perfect square (assuming it has one or after checking it has one) from $$3+2\sqrt{2}$$?

$$(a + b \sqrt{c})^2 = a^2 + b^2 c + 2 a b \sqrt{c}$$

So if you are given $$s + t \sqrt{c}$$ (with $$s, t, c$$ rational, $$t \ne 0$$ and $$c$$ not a square) and want to write it in this form, you want to find rationals $$a$$ and $$b$$ to solve the equations \eqalign{a^2 + b^2 c &= s\cr 2 a b &= t\cr} Since $$a=0$$ won't work, we can write $$b = t/(2a)$$ and the equation becomes $$a^2 + \frac{c t^2}{4a^2} = s$$ which we can solve for $$a^2$$ and then for $$a$$: $$a = \pm \frac{\sqrt{ 2\,s+2\,\sqrt {s^2-c{t}^{2}}}}{2}$$ That is, $$s^2 - c t^2$$ must be the square of a rational, and then $$2s + 2 \sqrt{s^2 - ct^2}$$ must be the square of a rational, and if so we get a solution with this $$a$$ and $$b = t/(2a)$$.

• I use to try $(a+\sqrt b)^2=a^2+b+2a\sqrt b$ first. – Yves Daoust Aug 4 at 15:25

Correction: $$(1+\sqrt 2)^2=3+\color{red}2\sqrt2$$

Generally:

$$(\alpha+\beta\sqrt x)^2=(\alpha^2+x\beta^2)+2\alpha\beta\sqrt x$$

Hence we can recursively solve using simultaneous equations:

$$\begin{cases}\alpha^2+2\beta^2=3 & \\ 2\alpha\beta=2 \end{cases}$$

Using the second to get $$\alpha=\frac 1\beta$$, we implant this into the first so:$$\frac{1}{\beta^2}+2\beta^2=3\implies2\beta^4-3\beta^2+1=0\implies(2\beta^2-1)(\beta^2-1)=0$$

This yields $$\beta=\pm1, \pm\frac{1}{\sqrt 2}$$ and thusly $$\alpha=\frac1\beta=\pm1,\pm\sqrt 2$$

The first pair gives us $$\pm(1+\sqrt 2)$$, and the seconnd pair gives $$\pm(\sqrt 2+\frac{1}{\sqrt2}\sqrt2)=\pm(1+\sqrt2)$$ which is the same result, achieved a different way.