# Can 2 identical square combine into a larger square? [closed]

I mean can twice the square of an integer be equal to another perfect square?

## closed as off-topic by Vidyanshu Mishra, user91500, Rohan, Shailesh, John BDec 31 '16 at 11:44

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• If by "number", you mean "integer" (or possibly "natural number"), the answer is either "no" or "yes, but only the trivial case $0^2=0^2+0^2$". This question is linked to the irrationality of $\sqrt{2}$. – πr8 Dec 31 '16 at 11:12

Assuming that $k,n\in\mathbb{N}$ satisfy

$$2k^2 = n^2 \tag1$$

then we have $($excluding the trivial case of $k=n=0)$

$$2k^2 = n^2 \implies 2=\left(\frac{k}{n}\right)^2\implies\sqrt{2} = \frac{k}{n}$$

This is a contradiction, as the square root of $2$ is irrational. Therefore, no $k, n\in\{\mathbb{N}\backslash 0\}$ exist that satisfy $(1)$.

• +1 for acknowledging the trivial case, and explicitly making the link to the irrationality of $\sqrt{2}$ – πr8 Dec 31 '16 at 11:14

Here's an alternative way using the Fundamental theorem of Arithemetic:
let $m^2=2n^2$. Then represent $n^2 = 2^{a_1}3^{a_2}\cdots$ in canonical form. Then, $a_1$ is even. Which means, $m^2 = 2^{a_1+1}3^{a_2}\cdots$. As, $a_1+1$ is odd, we have a contradiction.
Note: This method, too uses the irrationality of $\sqrt2$, but implicitly.

Let you have number n.

Then twice its square = $2n^2$ = 2 * n * n.

As you can see 2 as its one factor and 2 is not a perfect square. So twice the square of a number is not square.

• When you are down voting please comment the reason so that one can know the reason and try to improve. – Kanwaljit Singh Jan 1 '17 at 3:20

No. Because $a^2 + a^2 = 2a^2$.

But $\sqrt{2a^2}=\sqrt2\times a$.

Since $\sqrt2$ is irrational. $\sqrt2\times a$ is not an integer. $2a^2$ can never be perfect square.

As well mentioned in other answers, the sum of two squares will be $n^2+n^2=(\sqrt{2}n)^2$ which is not considered a perfect square as definition of perfect square is that :
Since $(\sqrt{2}n)$ is not an integer, so $(\sqrt{2}n)^2$ is not a perfect square.