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For this question, my way of proving it is:

write: $4|(2(\binom80a^8 + \binom81a^7b + \binom82a^6b^2... \binom88b^8) - (2a^8 +2b^8)$

since $2a^8 \ and\ \ 2b^8$ are eliminated, I just need to write $4|(\binom81a^7b + \binom82a^6b^2...\binom87ab^7)$ Because I know that for all $\binom8k$ where $0<k<8,\ k\in \mathbb{Z}$, $\binom8k$ is a multiple of 8, does this suffice as proof?

My second question is, is there a simpler (shorter) way around writing out $4|\binom81a^7b + \binom82a^6b^2...\binom87ab^7$?

Dharmesh Sujeeun
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  • What could be more simple, really? – aboat Mar 30 '16 at 03:29
  • part of my question is also whether this is deemed a correct proof? I need someone with higher authority on the subject than myself to confirm this for me. I have a quiz on modular arithmetic and cardinality tomorrow. – Dharmesh Sujeeun Mar 30 '16 at 03:32
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    "Correct proof" varies from prof to prof, anyway, usually on stackexchange you'll find people approving rather than dissapproving, because they get the idea. Your prof will most likely want you to follow his techniques rather than getting the idea. That's formal education. – aboat Mar 30 '16 at 03:34
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    There is a false assertion that $\binom{8}{k}$ is divisible by $8$ when $0\lt k\lt 8$. And there is a simple approach. – André Nicolas Mar 30 '16 at 03:45
  • Yeah, you're right. I was thinking permutations rather than combinations, which is why I messed up I think. – Dharmesh Sujeeun Mar 30 '16 at 04:07
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    @DharmeshSujeeun: Your proof will go through with minor modification, since all you need for mod $4$ is that $\binom{8}{k}$ is even whenever $1\lt k\lt 8$. For modulus $8$, we would need more, since $\binom{8}{4}=70$. – André Nicolas Mar 30 '16 at 04:11
  • Wep, j'comprends :-). À bientôt. – Dharmesh Sujeeun Mar 30 '16 at 04:14

1 Answers1

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There is a cases approach. If $a$ and $b$ are even, the result is clear. If $a$ and $b$ are both odd, then $(a+b)^8$ and $a^8+b^8$ are even, so the result holds.

If one of $a$ or $b$ is odd and the other even, then $(a+b)^8$ is odd, so $2(a+b)^8\equiv 2\pmod{4}$. But $a^8+b^8$ is also odd, so $2(a^8+b^8)\equiv 2\pmod{4}$.

One can strengthen the result. For example a very similar argument shows that the congruence holds modulo $8$. One can show with a bit more work that it holds modulo $32$.

André Nicolas
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