Can 2 identical square combine into a larger square? I mean can twice the  square of an integer be equal to another perfect square?
 A: 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 $(1)$.
A: 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.
A: 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.
A: 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.
A: The answer is NO. 
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 :

An integer that is square of another integer.

Since $(\sqrt{2}n)$ is not an integer, so  $(\sqrt{2}n)^2$ is not a perfect square.
