Prove that square of an integer is always of the form 3k or 3k+1 but never of the form 3k+2

I need help with this question.

[ Can someone tell that where can I find the solution manual for this book - Introduction to Number Theory By Ivan Niven And Hugh Montgomery - Some questions are really tough in this book]

Easy Math Editor

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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}`	$\boxed{123}$

Comments

$x\equiv0,1,2(mod3), x^{2}\equiv0,1,4(mod 3), \Rightarrow x^{2}\equiv0,1,1(mod3).$ As 0 and 1 occurs in the last step but 2 does not,Hence proved...

Kiran Patel - 7 years, 11 months ago

Use Euclid's division lemma.

Siddharth Kumar - 7 years, 11 months ago

its solution is present in the excursion

superman son - 7 years, 11 months ago

Induction can be used along with cases.

Ahmed Taha - 7 years, 11 months ago

You can divide the entire integer set into three parts: numbers that can be written as 3n , 3n+1 or 3n + 2. (3n)² = 9n² = 3 * 3n² = 3k. (3n + 1)² = 9n² + 6n + 1 = 3(3n² + 2n) + 1 = 3k + 1. And (3n + 2)² = 9n² + 12n + 3 + 1 = 3(3n² + 4n + 1) + 1 = 3k + 1.

Bruno Narciso - 7 years, 11 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}`	$\boxed{123}$