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I have a problem here hope you will help me.

Let a be an integer. If $a^2-3$ is odd then $3a+2$ is even.

I am asked to prove this by contradiction.

Example:
Proposition
For all integers $n$, if $n3 + 5$ is odd then n is even.

Proof.

Let $n$ be any integer and suppose, for the sake of contradiction, that $n 3 + 5$ and $n$ are both odd. In this case integers $j$ and $k$ exist such that $n 3 + 5 = 2k + 1$ and $n = 2j + 1$. Substituting for $n$ we have $$2k + 1 = n 3 + 5 2k + 1 = (2j + 1)3 + 5 2k + 1 = 8j 3 + 3(2j) 2 (1) + 3(2j)(1)2 + 13 + 5 2k = 8j 3 + 12j 2 + 6j + 5$$ We found $2k = 8j 3 + 12j 2 + 6j + 5$. Dividing by $2$ and rearranging we have $k − 4j 3 − 6j 2 − 3j = 5 2$ . This, however, is impossible: $5/2$ is a non-integer rational number, while $k − 4j 3 − 6j 2 − 3j$ is an integer by the closure properties for integers. Therefore, it must be the case that our assumption that when $n 3 + 5$ is odd then $n$ is odd is false, so $n$ must be even.

Gray_Rhino
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Nam do san
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    Welcome to MSE. Your question is phrased as an isolated problem, without any further information or context. This does not match many users' quality standards, so it may attract downvotes, or closed. To prevent that, please [edit] the question. This will help you recognise and resolve the issues. Concretely: please provide context, and include your work and thoughts on the problem. These changes can help in formulating more appropriate answers. – José Carlos Santos Nov 16 '21 at 02:49

1 Answers1

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There is no need for a proof by contradiction. $a^2 - 3$ is odd $\implies a^2$ is even $\implies a$ is even $\implies 3a + 2$ is even. $\blacksquare$

Edit: A proof by contradiction. If possible, suppose $3a + 2$ is odd. As $2$ is even, $3a$ must be odd. So, $a$ must be odd. This implies that $a^2$ is odd, and so $a^2 - 3$ is even (difference of two odd numbers is even). This contradicts our hypothesis that $a^2 - 3$ is odd. So, $3a + 2$ must be even. $\blacksquare$

stoic-santiago
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