Help on proving this statement...

**free_to_fly** · July 20th 2008, 11:35 AM

Prove that if n=2 (mod 3) then it cannot be a square.

No idea where to start with this one. Any pointers please?

**Moo** · July 20th 2008, 11:44 AM

Originally Posted by free_to_fly

Prove that if n=2 (mod 3) then it cannot be a square.

No idea where to start with this one. Any pointers please?

Hmm a table of congruences would work.

Assume that $n=k^2$

$\text{If } k=0 \bmod 3 \text{ then } n=k^2=0 \bmod 3$

$\text{If } k=1 \bmod 3 \text{ then } n=k^2=1 \bmod 3$

$\text{If } k=2 \bmod 3 \text{ then } n=k^2=4 \bmod 3=1 \bmod 3$

And these are all the possibilities. You can see that none of the results are $2 \bmod 3$

**ThePerfectHacker** · July 20th 2008, 11:47 AM

Originally Posted by free_to_fly

Prove that if n=2 (mod 3) then it cannot be a square.

No idea where to start with this one. Any pointers please?

Here is an interesting fact. The only (odd) primes that have 2 as a square are of the form $8k\pm 1$ . (Since $p=3$ does not have this form this form it is not a square of $2$ . This is not meant to be a proof of what you asked, just an interesting fact).

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