# How did Hardy make this step in proving two surds are dissimilar if they are non-perfect square integers with no common factor

Overview

I am attempting to replicate a proof Hardy provided in his book - A Course of Pure Mathematics, yet am having trouble with one of the steps. I was wondering if someone can explain how he made a step in the proof.

Statement

If $M$ and $N$ are integers which have no common factor, and neither of which is a perfect square, $\sqrt{M}$ and $\sqrt{N}$ are dissimilar surds.

His Proof up to the Point of Confusion

Suppose that $\sqrt{M}$ and $\sqrt{N}$ are similar surds. Then we can instead write them as: $\sqrt{M}=\dfrac{p}{q} \sqrt{\dfrac{t}{u}}$ and $\sqrt{N}=\dfrac{r}{s} \sqrt{\dfrac{t}{u}}$

Then $\sqrt{MN}$ is evidently rational, and therefore (from a previous example) integral.

The example he is referring to

An algebraic equation,

$x^n+p_1 x^{n-1} +p_2 x^{n-2}+...+p_n=0$

with integral coefficients, cannot have a rational but non-integral root.

My Question

How was Hardy able to determine that $\sqrt{MN}$ was integral from that example he was referring to?

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We want to show that if $C$ is an integer, and $\sqrt{C}$ is rational, then $\sqrt{C}$ is an integer.

So we want to show that any rational solution of the equation $x^2-C=0$ is actually an integer.

Let $\frac{a}{b}$ be a rational solution of the equation. We may without lloss of generality assume that $\frac{a}{b}$ is in lowest terms, that is, that no integer $\gt 1$ is a common divisor of $a$ and $b$. We may also without loss of generality assume that $b$ is positive.

Substituting in the equation, we find that $\left(\frac{a}{b}\right)^2-C=0$.

Multiply through by $b$. We find that $$a^2=b^2C.$$ Now let $p$ be any prime divisor of $a^2$. Since $a^2=b^2 C$, the prime $p$ must divide $a^2$, so it must divide $a$. This is impossible, since $\frac{a}{b}$ is in lowest terms.

We conclude that $b$ has no prime divisors, meaning that $b=1$. It follows that $C=a^2$, a perfect square, so $\sqrt{C}=a$, an integer.

Remark: The argument we used for the polynomial equation $x^2-C=0$ can, without significant modification, be used to prove the theorem about $$x^n+p_1 x^{n-1} +p_2 x^{n-2}+\cdots+p_n=0$$ that you quoted. Once it has been proved, it can be applied to our particular equation $x^2-C=0$.

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Because $\:x = \sqrt{MN}\:$ is a root of $\:x^2 - MN = 0\:$ so, being rational, it is integral, by said theorem (which is known as the rational root test).

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