Debilitating Perfect Squares

Number Theory Level pending

What is the number of pairs of integers such that their perfect squares are in the ratio of 2 to 1?

The answer is 0.

1 solution

Harshit Singhania
Jul 2, 2015

Let the perfect squares be $p^{2}$ and $q^{2}$ then $\frac{p^{2}}{q^{2}} = 2 >> \frac{p}{q} = \sqrt{2}$ as $\sqrt{2}$ is an irrational number so it cannot be represented in the form of $\frac{p}{q}$ where p and q are integers hence the answer is 0

It may also be $-\frac{p}{q}=\sqrt{2}$ , but the result still follows. However, this is circular reasoning, because the most common proof of the irrationality of $\sqrt{2}$ proves $\frac{p^2}{q^2}=2$ has no integer solutions.

$p^2=2q^2$ is impossible for $p,q\neq 0$ , because LHS has an even power of $2$ and RHS has an odd power of $2$ . Alternatively, $p=2k$ , so $2k^2=q^2$ , so $q=2m$ , so $k^2=2m^2$ . By infinite descent $p=q=0$ .

mathh mathh - 5 years, 11 months ago

Debilitating Perfect Squares

The answer is 0.

1 solution

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