2
$\begingroup$

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

$\endgroup$

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

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Vidyanshu Mishra, user91500, Rohan, Shailesh, John B
If this question can be reworded to fit the rules in the help center, please edit the question.

  • 1
    $\begingroup$ 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 $\sqrt{2}$. $\endgroup$ – πr8 Dec 31 '16 at 11:12
3
$\begingroup$

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 $(1)$.

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

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.

$\endgroup$
1
$\begingroup$

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.

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

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.

$\endgroup$
0
$\begingroup$

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.

$\endgroup$

Not the answer you're looking for? Browse other questions tagged or ask your own question.