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Let n, m be integers. Prove that if n ≡ 1( mod 2) and m ≡ 3( mod 4), then n^2 +m ≡ 0(mod4)

My thought process is that n=3 and m=7 so n^2 would be 9+7 but then i do not believe that would equal 0(mod4) which i think would equal 4.

Any direction would help on where to start or how to attack this problem.

Antonio Vargas
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Cole
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2 Answers2

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Hint: Consider two cases:

$n\equiv 1\text{mod}4$

or

$n\equiv 3\text{mod}4$

What can you say about $n^2\text{mod}4$ in either of those cases?

JMoravitz
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  • The only thing I can think of is that since it is squared it would be even – Cole Sep 17 '15 at 04:44
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    @Cole If $n\equiv 1\text{mod}2$ then $n$ is odd. If $n\equiv 1\text{mod}2$ then either $n\equiv 1\text{mod}4$ or $n\equiv 3\text{mod}4$. An odd number squared is always odd. Why would you think it would be even? – JMoravitz Sep 17 '15 at 04:47
  • Ya you are right sorry. it's been a really long day. I screwed up. – Cole Sep 17 '15 at 04:48
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$n \equiv 1( \mod 2)$ means $\exists a \in \mathbb{Z}$ such that $n = 2\cdot a + 1$, so $$ n^2 = (2\cdot a + 1)^2 = 4a^2 + 4a +1.$$

So you can calculate $n^2 +m \mod 4$ now.

corindo
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