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How would you prove $2\mid (n^4-3)\iff 4\mid (n^2 + 3)$. Can you use induction on both at same time? Can I just do it directly somehow?

Dave
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andemw01
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3 Answers3

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$4 \mid (n^2+3) \implies n$ is odd $\implies n^4$ is odd $\implies n^4 -3$ is even $\implies 2 \mid (n^4 - 3)$.

$2 \mid (n^4 - 3) \implies n$ is odd $\implies n = 2k + 1$ for some $k \in \mathbb{Z} \implies n^2 = 4k^2 + 4k + 1 \implies n^2 + 3 = 4k^2 + 4k +4 \implies 4 \mid (n^2 + 3)$.

jgsmath
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    Just as a tip, you can use "\mid" instead of "|" to show 'divides'. e.g. $2\mid 4$ as opposed to $2|4$. – Dave Jun 29 '17 at 01:30
  • This was very useful....I used it in a very round about way with the fact n must be odd both ways since iff statement...Thank You. – andemw01 Jun 29 '17 at 02:48
  • @Dave: Thanks for the tip.. It will be useful to me in the future. :-) – jgsmath Jun 29 '17 at 14:48
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$$2\mid n^4-3 \iff 2\mid(n^2+1)(n+1)(n-1)-2$$

$$\iff 2\mid n+1 \text{ and } 2\mid n-1 \iff 4\mid n^2-1 \iff 4\mid n^2+3.$$

dromastyx
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$2 \mid (n^4 - 3)$ iff $2\cdot 2 \mid 2\cdot(n^4 - 3)$. So it is equivalent to proving

$$2n^4 - 6 \equiv 0 \text{ iff } n^2 + 3 \equiv 0 \pmod 4$$

And there are only 4 cases you need to check ($0 \le n < 4$).

DanielV
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