I'm going back to the magazine from the previous post.
Starting with this issue, 1 / 1979, a new column appears, PREPARATORY PROBLEMS FOR IMO (see pages 32-33). At that time, I was in my last year of High School and I decided (at the suggestion of a high school classmate, Mariana DIACONESCU) to enroll at the Faculty of Mathematics.
I think that during the summer vacation after my high school diploma, after I passed the college entrance exam, I tried my hand at these problems. Problem O : 1 was a particular case of the article published in the same journal, on pages 5-7, and I imitated the solution there. Problem O : 2 I didn't like at the time, so I moved on to O : 3.
"O : 3. Let $p$ be a prime number, $p>2$ . For each number $k$ from
$1\; to\; p-1$ , we denote by $a_k$ the remainder of the division of $k^p\;by\;p^2$ .
Let it be shown as
$$a_1+a_2+a_3+\dots+a_{p-1}=\frac{p^3-p^2}{2}$$
{Taken from : }( Kvant Magazine, issue 2 / 1978)
ANSWER CiP
$$a_k+a_{p-k}=p^2\;\;\;\;\;,\;\;k=1,\;2,\;\dots,\;p-1 \tag {1}$$
Solution CiP
According to division with remainder we can write :
$k^p=A_k\cdot p^2+a_k\;\;\;,\;\;0<a_k<p^2\;\;\;\;,\;\;\;A_k\in\mathbb{N}\;\;,\;\;k=1,\;2,\;\dots\;, p-1 \tag{2}$
Since the number $p$ is prime, we have $p\not \mid k^p$ so $a_k$ cannot be zero.
We can also write :
$(p-k)^p=A_{p-k}\cdot p^2+a_{p-k}\;\;\;,\;\;0<a_{p-k}<p^2 \tag{3}$
Adding the equations (2) and (3) we have :
$k^p+(p-k)^p=B_k\cdot p^2+(a_k+a_{p-k}) \tag{4}$
where $B_k=A_k+A_{p-k}\in\mathbb{N}$. But, according to the binomial theorem :
$(p-k)^p=p^p-C_p^1\cdot p^{p-1}\cdot k+C_p^2\cdot p^{p-2}\cdot k^2-\dots -C_p^{p-2}\cdot p^2\cdot k^{p-2}+C_p^{p-1}\cdot p\cdot k^{p-1}-k^p$
We denoted the binomial coefficients with $C_p^k=\frac{p!}{k!\cdot (n-k)!}$, which in latex is denoted $_p^k\textrm{C}$ , and in algebra $\binom{p}{k}$.
Since $C_p^{p-1}=p$ , we have from here
$k^p+(p-k)^p=\left [p^{p-2}-C_p^1p^{p-3}k+C_p^2p^{p-4}k^2-\dots -C_p^{p-2}k^{p-2}+k^{p-1} \right ]\cdot p^2$
so the left side of the equation (4) is divided by $p^2$ and then on the right side we must have $p^2\mid (a_k+a_{p-k})$. So
$a_k+a_{p-k}=D\cdot p^2\;\;,\;\;D\in\mathbb{N} \tag{5}$
But from the inequalities in (2) and (3) it follows $0<a_k+a_{p-k}<2\cdot p^2$ . So in (5) we must have $D=1$ , that is, precisely the formula (1).
Adding the $p-1$ relations (1) we have, written from right to left :
$$(p-1)\cdot p^2=\sum_{k=1}^{p-1}(a_k+a_{p-k})=\sum_{k=1}^{p-1}a_k+\sum_{p-k=1}^{p-k=p-1}a_{p-k}=2\cdot \sum_{k=1}^{p-1}a_k$$
The formula in the problem is proven.
$\blacksquare\;\;\;\blacksquare$
For example :
$a_1=remainder\;1^7:49\;\;=1$
$a_2=remainder\;2^7:49\;\;=30$
$a_3=remainder\;3^7:49\;\;=31$
$a_4=remainder\;4^7:49\;\;=18$
$a_5=remainder\;5^7:49\;\;=19$
$a_6=remainder\;6^7:49\;\;=48$
and it is seen that $a_1+a_6=a_2+a_5=a_3+a_4=49$.
$\square$
REMARK CiP This problem gave me a lot of satisfaction when I solved it. At first I tried different ways to approach it, without success. It was only after I did a numerical simulation with a pocket calculator (as in the example presented above) that I noticed the fulfillment of the relation (1) between the residues. I was delighted with the discovery I made. I was going to demonstrate it...The idea of showing that we have D equal to 1 in the relation (5) was first encountered in Sierpinski (in Russian, in Romanian - at that time I could easily and long-term retain what I read). Today, it took me a few days to reconstruct the demonstration from back then.
And, let's report another error. The problem had the number M442 and was published in issue 5 / 1977 (page 20). In issue 2 / 1978 the solution was published (page 28).
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