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Do we have any possibility that we get ,

$ p^2 = 2q^2 $, $ \forall p,q \in {N} $

I just reading the properties of Square Numbers on [ Properties of Square numbers ]

I found that property no.7: There are $n$ natural numbers $p$ and $q$ such that $p^2 = 2q^2$. I tried but did not get any pair of numbers.

I tried some examples:

  1. If we have $12^2=144=2(72)$, and $72$ is not perfect square

  2. $10^2=100=2(50)$, and $50$ is not perfect square

  3. $ 22^2=484=2(242)$, and $242$ is not a perfect square

  4. $ 16^2=256=2(128)$, $128$ is not perfect square.

Do we have any possibility with $p,q\in\mathbb{N}$ such that $p^2=2q^2$ holds?

Bill Dubuque
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Atique Ahmed
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    We can't have $p^2=2q^2$ with $p,q\in\mathbb N$, because otherwise $\pm\sqrt2=p/q$, which contradicts the fact that $\sqrt2$ is irrational – J. W. Tanner Apr 23 '21 at 21:06
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    @lulu $p^2=2q^2 \to (1)$ , Sir its mean this equation a name as equation no.1 so it doesn't need to write $p^2=2q^2 $ again and again. If u think thats something not understandable . i will edit the question – Atique Ahmed Apr 23 '21 at 21:13
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    "There are $n$ natural numbers $p$ and $q$ such that $p^2=2q^2$" . I checked the website that you referenced, which does actually give the above assertion. This is a typo on the website. The assertion should read: "There are no natural numbers $p$ and $q$ such that $p^2=2q^2$" . – user2661923 Apr 23 '21 at 21:15
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    @user2661923 Okay Sir I got it.. – Atique Ahmed Apr 23 '21 at 21:18

2 Answers2

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One way to see that this is not possible is, as was mentioned in the comments, it contradicts the irrationality of $\sqrt 2$. In fact, this is usually used to prove that $\sqrt 2$ is irrational.

To see that this is impossible, suppose by way of contradiction that such a $p$ and $q$ exist. We can assume that $p$ and $q$ are coprime. Then, $p^2 = 2q^2$ implies that $p^2$ is even, and thus $p$ is also even. But, then $4$ divides $p^2$ and thus also divides $2q^2$. This implies that $q^2$ is even and so $q$ is even. This contradicts the assumption that $p$ and $q$ are coprime.

Gary Moon
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$q^2$ is a square for definition, so the exponent of $2$ present in the prime factorization of $q^2$, as well as the exponent of every other prime, must be even. Doubling $q^2$ changes that exponent from even to odd. But if the prime factorization of $2q^2$ has $2$ raised to an odd exponent, it can't be a square, since no square has an odd exponent in its prime factorization.

Claudio_DiBiase
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