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Let $a$ be an integer which is not divisible by 2 and 3. Prove that 24 divides $a^2-1$.

This, $a$ can be written as $a=2x+1$ or $a=3z+r$ where $r=1,2$ and $x$ and $z$ are integers. This $a^2-1= (2x+1)^2-1$ or $(3z+r)^2-1$

Taking first case : $4x^2+4x$

But, how to prove that it is divisible by 24? Is my method correct? Please give me some hints on how to do this.

3 Answers3

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Put $a=6k+r$ then $r=1$ or $r=-1$. Then $a^2-1=36k^2\pm 12k=12(3k+1)k$. If $k$ even this is divisible by $24$ and also when $k$ is odd.

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Hint: In this kind of divisibility problem it is sometimes useful to work with different primes separately - consider divisibility by $8$ and $3$ and then combine the answers, as you are doing. $4x^2+4x$ is obviously divisible by $4$ - can you prove it is divisible by $8$?

Alternative: $a^2-1=(a+1)(a-1)$ since $a$ is odd, both these factors are even - add some other observations about consecutive even integers, and about three consecutive integers.

Another alternative: show that the divisibility criteria mean that $a=6b\pm 1$ for $b$ an integer.

Mark Bennet
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Hint : Since the number a is not divisible by 2 and 3 consider a number 6b ( where b is some integer ) which is divisible by both 2 and 3. So adding 1 or subtracting 1 will give you number like a.

After substituting the value of a as $ 6b+1$ or $6b-1$ in $a^2-1$ will give you $36b^2+12b$ or $36b^2-12b$ . Now resolve this into factors and divide it by 24 you will observe that it is clearly divisible by 24.

Jasser
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