Let $a$ be an integer that is not divisible by either $2$ or $3$. Prove that $a^2 − 1$ is divisible by $24$.

Question

Let $a$ be an integer that is not divisible by either $2$ or $3$. Prove that $a^2 − 1$ is divisible by $24$.

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I'm using congruence for this. An idea is to use the division algorithm to get some more useful value for $a$. That is, $a$ equals either $6k+1$ or $6k+5$, because it is not divisible by either 2 or 3 (so, e.g., we can't have $a=6k+2$ since that is divisible by 2, etc...)

The next step would be to square this and use the properties of congruence arithmetics to should that the remainder of $(a^2 − 1)/24$ is $0$.

My problem is I'm not making much progress when I substitute $a=6k+1$ or $6k+5$ into $a^2 − 1$.

Any help? Thanks.

Answer 1

Just see that $ (6k \pm 1)^2 - 1 = 36k^2 \pm 12k = 24k^2 + 12k(k \pm 1) $.

Answer 2

Note that $a^2-1 = (a-1)(a+1)$

Consider the three consecutive integers, $\{a-1, a, a+1\}$.

One of them is a multiple of $3$ but that one is not $a$ so it is one of $a-1$ or $a+1$.

On the other hand, since $a$ is odd, both $a-1$ and $a+1$ are even and one of them is a multiple of $4$

Thus the product $(a-1)(a+1)$ is a multiple of $3$ and $8$ which makes it a multiple of $24$

Answer 3

An integer which is divisible neither by $2$ nor by $3$ is congruent to $\pm1,\pm 5,\pm 7,\pm 11 \bmod 24$. The square of each of these numbers is congruent to $1\bmod24$.