Is it as easy as it looks?

Prove that the square of any positive integer is in the form of either $3m+1$ or $3m$ for some integer $m$.

$n^{2}$ = $3m+1$

$n^{2}$ = $3m$

I couldnt get this, so decided to post it to see if anyone knows how it can be done Post your proofs below please :)

#NumberTheory #Is_It_aS_eaSy_as_it_looks_? #Simple_proofs #N_Squares

Easy Math Editor

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Comments

So this question could also be phrased, show that n^2 is 0 or 1 modulo 3.

The great thing about modulo is that you can multiply things. So we only need to try n being 0, 1, 2 modulo 3.

2^2 = 4 = 1 mod 3 The other two are trivial and left as an exercise for the reader.

Nathan Edwards - 6 years, 10 months ago

Thknx

Chinmay Raut - 6 years, 10 months ago

FIRST- a^2={3q}^2 a^2=9q^2 a^2=3(3q^2) a^2=3m,where m=3q^2 SECOND- a=3q+1 a^2={3q+1}^2 a^2=9q^2+1+2(3q*1) a^2=9q^2+1+6q a^2=3(3q^2+2q)+1 a^2=3m+1,where m=3q^2+2q

Anmol Singh - 6 years, 10 months ago

Thnks

Chinmay Raut - 6 years, 10 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$