# 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]

Note by Kishlaya Jaiswal
4 years, 11 months ago

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

- 4 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.

- 4 years, 11 months ago

Induction can be used along with cases.

- 4 years, 11 months ago

its solution is present in the excursion

- 4 years, 11 months ago

Use Euclid's division lemma.

- 4 years, 11 months ago