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I tried to expand this sum using the formula but still can't observe why it can be divided by... I mean eventually, I'll get $(p-1)(2p-1)$ must be divided by $6$, but why?

J. W. Tanner
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WWMASK
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    Does this answer your question? Showing that $1^k+2^k + \dots + n^k$ is divisible by $n(n+1)\over 2$. Now take $n=p-1$. – Dietrich Burde Jan 29 '20 at 14:42
  • It will help if you show the formula and how you used it. You may have made a mistake somewhere, but it's hard to tell without seeing what you tried. – Barry Cipra Jan 29 '20 at 14:43
  • Another way to see it can be to see that $1^2+2^2+...+(p-1)^2=(1+2+...+(p-1))^2-2(1\cdot 2+1\cdot 3+...+(p-2)\cdot (p-1))$. Now, by Fermat's theorem $1,2,3,...p-1$ all are solutions of $x^{p-1}-1=0\pmod{p}$. Note that $1+2+...+(p-1)$ and $1\cdot 2+1\cdot 3+...+(p-2)\cdot (p-1)$ are the coefficients, up to sign, of $x^{p-1}-1$ of the terms of degree $p-2$ and $p-3$, respectively. Therefore, they are both divisible by $p$. This idea is not faster for this particular problem, but it is a useful one in many others involving symmetric sums. – OscarRascal Jan 29 '20 at 14:47
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    You should have mentioned that $p$ is a prime, otherwise this is not true. – Sil Jan 29 '20 at 14:53

2 Answers2

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$1^2+2^2+3^2+\cdots+(p-1)^2=\dfrac{(p-1)p(2p-1)}6,$

so $p$ divides $6(1^2+2^2+3^2+\cdots+(p-1)^2)$.

But, if $p\ge5$, then $p$ doesn't divide $6$.

Now use Euclid's lemma.

J. W. Tanner
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In order to complete your proof , note that for a prime number $p\ge 5$ we have $ (p-1)(2p-1)=2p^2-3p+1$, which is clearly an even number.

In order to show that it is also a multiple of $3,$ we show that $2p^2+1$ is a multiple of $3.$

Since $p$ is prime, we have $p=3k\pm 1$; thus $2p^2+1=2(3k\pm 1)^2+1$ is a multiple of $3$, and the proof is complete.

J. W. Tanner
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Mohammad Riazi-Kermani
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