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The expression $a^3+11a-6$ is always divisible by 6.

I came across it while programming and ran it in a function in a calculator.

Upon trying to solve it by myself, I found $a^3+11a-6$ is always even, as when $a^3$ is odd so is $11a$ and when $a^3$ is even so is $11a$, and odd+odd=even and even+even=even.

Now, I tried to find if the number is divisible by $3$, but I couldn't. Can anyone help me out? Why should this expression always be divisible by $6$, we know it is but why?

And is there a way to find if an expression is divisible by a certain number. Thanks for any help.

Jyrki Lahtonen
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Samip Gyawali
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  • This has nothing whatsoever to do with [tag:division-algebras]. Please check the meaning of an unfamiliar tag before using it. – Jyrki Lahtonen Jun 02 '21 at 13:33

2 Answers2

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Actually, $a^3-a$ is always divisible by $6$, so that also $$ a^3-a+(12a-6)=a^3+11a-6 $$ is always divisible by $6$.

Reference:

Proving divisibility of $a^3 - a$ by $6$

Dietrich Burde
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HINT:

$a^3+11a-6=a^3-a+12a-6=a(a+1)(a-1)+6(2a-1)$

What can we say about the product of 3 consecutive numbers?

Star Bright
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