AoPS Academy
on the board. Every minute Ivan makes next operation: takes any number 
written on the board, except ( Can be same numbers on the board). After some time on the board there are 
numbers. is it possible?
be real numbers satisfying 
Prove that 
and determine all cases of equality.
be a positive integer and let 
be distinct integers in the set 
such that divides 
for 
. Prove that does not divide 
on 
without derivative.
there exists a positive integer 
such that![\[1^n+2^{n-1}+3^{n-2}+\cdots+n^1\equiv 2020\pmod{p}.\]](http://latex.artofproblemsolving.com/b/6/9/b6942b4c676c2ed55aab4fc85ae73c86b47998d3.png)
Robin Son
.![$f(x+(p-1)) \equiv f(x) + [(x+1)^{p-1} + (x+2)^{p-2}+\cdots+(x+p-1)^1] \pmod p$](http://latex.artofproblemsolving.com/5/f/5/5f5aca146ae333b01ef9ed68472abf41ea4b9e21.png)
.
and using FLT. 
.
times gives![\begin{align*}
f(x+p(p-1)) &\equiv f(x) + [(x+1)^{p-1} + \cdots + (x+p-1)^1] + [(x+p)^{p-1} +\cdots + (x+2(p-1))^1 ] \\
&\qquad \ \ \ \ \ + \cdots \\
&\qquad \ \ \ \ \ + [(x+1+(p-1)^2)^{p-1} + \cdots + (x+p(p-1))^1 ] \\
&\equiv f(x) + [(x+1)^{p-1} + (x+p)^{p-1} + \cdots + (x+1+(p-1)^2)^{p-1}] \\
&\qquad \ \ \ \ \ + \cdots \\
&\qquad \ \ \ \ \ + [(x+p-1)^1 + (x+2p-2)^1 + \cdots + (x+p(p-1))^1 ] \pmod p
\end{align*}](http://latex.artofproblemsolving.com/d/e/c/decb95465f73eb45531488a0c2fd24a53c5fcb4c.png)
where the last step follows by regrouping. Call 
the mess after 
above in mod . We aim to show that 
. Each of the 
bracketed terms above is of the form![\[ (x+k)^{p-k} + (x+k+(p-1))^{p-k} + \cdots + (x+k+(p-1)^2)^{p-k}\]](http://latex.artofproblemsolving.com/e/e/c/eec858a676a86a8e9d02c1d23a4e2304ea5ec103.png)
for 
. But since 
has each term increasing by , this forms an entire residue class mod . Therefore, each bracketed term is actually just![\[ 0^{p-k} + 1^{p-k} + \cdots + (p-1)^{p-k}. \]](http://latex.artofproblemsolving.com/b/8/6/b86044ce8961261049d31ced0bf63a98ddaf1c25.png)
The entire sum is therefore![\begin{align*}
S &= \sum_{k=1}^{p-1} [0^{p-k} + \cdots + (p-1)^{p-k}] \\
&\equiv \sum_{\ell=0}^{p-1} [\ell^1 + \ell^2 +\cdots + \ell^{p-1}] \\
&\equiv \left[ \sum_{\substack{\ell=0 \\ \ell \neq 1}}^{p-1} \left[ \frac{\ell^p - \ell}{\ell-1} \right] \right] + [1^1+1^2+\cdots + 1^{p-1}] \\
&\equiv 0 + (p-1) \equiv -1 \pmod{p}
\end{align*}](http://latex.artofproblemsolving.com/2/f/e/2fe4a9674f542dc653c9a853ddc8575b449df9e6.png)
as desired. Note that the second step above followed by regrouping, the third by geometric series, and the last by FLT. This proves the claim. 
is in the range of 
, as we can iterate the claim to get all residues. Therefore, 
is also in the range of , and we are done.
works.
works.
)
works.
. Consider 
for some 
, motivated by the fact that some things here are periodic with period (bases) and (exponents). Observe that
Now, notice that, as all exponents are positive, and 
for any 
,
as 
will range over all residues mod as 
ranges over all residues mod . Now, for 
the sum will be 
, but for 
this sum will be , so the whole thing is congruent to 
so our original sum is congruent to 
. Now set 
.
. It is well known that 
if 
, and 
if 
. Thus, we can compute 
. But also note 
, so we're done.
. It is well known that if , and if . Thus, we can compute . But also note , so we're done.
summation
is trivial. I claim that 
works, where 
. Denote the sum on the LHS by . Take mod , noting 
: ![\[S=1^n+2^{n-1}+\cdots +n^1=\sum_{k=1}^n (n+k-1)^k\equiv \sum_{k=1}^n (-k)^k \pmod p\]](http://latex.artofproblemsolving.com/2/a/a/2aafe6e41652a811941c5aed886feaaa2abcddbe.png)
We can ignore all multiples of since they vanish. Using FLT, we get ![\[S\equiv \sum_{k=1}^n (-k)^k=\sum_{k=1}^{p-1}(-k)^k+(-k-p)^{k+p}+\cdots +\left(-k-p\left(\frac{n+1}{p}-1\right)\right)^{k+p\left(\frac{n+1}{p}-1\right)}\]](http://latex.artofproblemsolving.com/f/2/1/f21bf1dfc881c8fe908d6f867a81431c1be65940.png)
![\[\implies S\equiv \sum_{k=1}^{p-1} (-k)^k+(-k)^{k+1}+\cdots +(-k)^{k+\left(\frac{n+1}{p}-1\right)} \pmod p\]](http://latex.artofproblemsolving.com/f/3/c/f3ce9f6fb0072650dc36a3110196f73d1a791961.png)
Note that 
is a set of 
consecutive integers. Thus, under 
, the exponents reduce to copies of the complete residue system mod : ![\[\implies S\equiv a\cdot \sum_{k=1}^{p-1} (-k)^1+(-k)^2+\cdots +(-k)^{p-1}\equiv a\cdot \sum_{k=1}^{p-1} k^{1}+k^2+\cdots +k^{p-1} \pmod p\]](http://latex.artofproblemsolving.com/4/f/c/4fca510e223174d317b316879f4e7c624b167270.png)
where we get the last equality from the fact that 
is a complete residue system mod excluding 
. Rearranging, we get ![\[S\equiv a\cdot \sum_{k=1}^{p-1} 1^k+2^k+\cdots +(p-1)^k\pmod p\]](http://latex.artofproblemsolving.com/2/8/8/288261fadc1cc7956016626bbbf4815515e2aa0b.png)
However, the summand is well known to be mod if does not divide 
, and mod if divides . Thus, ![\[S\equiv -a\equiv 2020\pmod p\]](http://latex.artofproblemsolving.com/5/c/d/5cd498baa0159fa587165cad6aca87de57d295a2.png)
as desired.
but without the minus one, so if anyone could proofread and let me know if I went wrong somewhere, it would be much appreciated 
. times gives Actually it's kind of a backward procedure (as I think) means
as 
then by applying the lemma we write it as![$f(x+p(p-1)) \equiv f(x+(p-1)^2)+[(x+1+(p-1)^2)^{p-1}+(x+2+(p-1)^2)^{p-2} + \dots + \underbrace{(x+(p-1)+(p-1)^2)^{1}}_{(x+p(p-1))^{1}}] \pmod p$](http://latex.artofproblemsolving.com/d/b/c/dbcea9466f783f006f0f737ef4963b3fa2c5e8c4.png)
as 
and thus we can again apply the lemma and we continue with the process... 
works. The main motivation is just playing around with the sum and some wishful thinking.
for 
Then note that the sum is equivalent to
to get
and 
This motivates setting 
so that there are no numbers between to 
inclusive that lie in the second category.
Since all of the terms divisible by disappear, we're left with
is a permutation of
since exactly one of the sums inside is congruent to 
or
Thus 
works, so
suffices, as desired.
and integers 
, ![\[1^k+2^k+\cdots+(p-1)^k\equiv\begin{cases}-1&\text{if }p-1\mid k\\0&\text{otherwise.}\end{cases}\]](http://latex.artofproblemsolving.com/3/4/5/345388a57333ea72dd78c75a2d45070db0dcfddf.png)
, end by Fermat's little theorem. Else let 
be a primitive root, and write ![\[1^k+\cdots+(p-1)^k\equiv1+g^k+\cdots+g^{(p-2)k}\equiv\frac{g^{(p-1)k}-1}{g^k-1}\equiv0\pmod p.\]](http://latex.artofproblemsolving.com/2/6/b/26b33ac7e21c8736a993dd9d540cc37261b3717d.png)
![\[f(n):=1^n+2^{n-1}+\cdots+n^1.\]](http://latex.artofproblemsolving.com/6/2/6/6268b6c0687885c726624f0a0fc7d57c40fe473f.png)
The main claim for this problem is this:
.![\begin{align*} f(n+p(p-1))&=\left[1^{p(p-1)+n}+\cdots+n^{p(p-1)+1}\right]+\left[(n+1)^{p(p-1)}+\cdots+(n+p(p-1))^1\right]\\ &\equiv\left(1^n+\cdots+n^1\right)\\ &\qquad+\left[(n+1)^{p(p-1)}+n^{(p-1)^2}+(n-1)^{(p-2)(p-1)}+\cdots\right]\\ &\qquad+\left[(n+2)^{p(p-1)-1}+(n+1)^{(p-1)^2-1}+n^{(p-2)(p-1)-1}+\cdots\right]\\ &\qquad\;\vdots\\ &\equiv f(n)+\sum_{k=0}^{p-2}\left(1^k+\cdots+p^k\right)\\ &\equiv f(n)-1\pmod p. \end{align*}](http://latex.artofproblemsolving.com/f/8/5/f85895efcf786ea7fa3a4b06298b31938b0a276f.png)
is surjective modulo .
exponent equation with a lot of fixed parameters and one which appears on the exponents and bases, then 
We first count the value of the expression 
, that![\[ \sum_{i=1}^{p(p-1)}i^{[p(p-1) + 1]-i} \equiv -1 \pmod p \]](http://latex.artofproblemsolving.com/f/e/8/fe846c1afd248bb8584c838d2d27fb4e3a535b48.png)
We categorize the 
exponents according to their bases . For 
, we have
with the third congruence to be true as the set 
and the fifth because 
and 
.
, it's easy to see that the expression![\[ i^{p(p-1)+1-i} + {(i+p)}^{p(p-2)+1-i} + \ldots + {(i+p(p-2)}^{p+1-i} \]](http://latex.artofproblemsolving.com/c/0/7/c07c89871dcd8a1b5b83d0241124ca00f48b80ee.png)
is equal to 
. Summing them all yields the desired congruence. 
Now let 
, with 
. We will prove 
works.
terms into 
chunks of perfect powers --- all of which are equivalent to the expression in 
above. In more detail, the term 
will appear times, in the form of![\[ {(i)}^{q \cdot p(p-1)+1-i}, {\textcolor{red}{(}i+p(p-1)\textcolor{red}{)}}^{\textcolor{magenta}{(}(q-1) \cdot p(p-1)+1-i\textcolor{magenta}{)}}, \dots, {\textcolor{red}{(}i+(q-1)\cdot p(p-1)\textcolor{red}{)}}^{\textcolor{magenta}{(}p(p-1)+1-i\textcolor{magenta}{)}}
\]](http://latex.artofproblemsolving.com/2/4/9/24953f052f3cb4646e2d8de872bb3ca16c110148.png)
Then we are done, because the whole thing is congruent to 
.. The original sum becomes :![\[1^{kp(p-1)} + 1^{kp(p-1) - p} + \cdots + 1^{kp(p-1) - p(p-2)} \quad + \]](http://latex.artofproblemsolving.com/0/9/0/090af67c18898adf0ea8d42c7abe9957e6b74a84.png)
![\[2^{kp(p-1) - 1} + 2^{kp(p-1) - (p+1)} + \cdots + 2^{kp(p-1)- (p(p-2) + 1)} \quad + \]](http://latex.artofproblemsolving.com/a/a/c/aac0d7df0b3e3098c1099a17a0c01b89065afc81.png)
![\[.\]](http://latex.artofproblemsolving.com/f/e/d/fed957b50425f14f13a98df5a86687676bbde0b4.png)
![\[+ p-1^{kp(p-1) - (p-2)} + (p-1)^{kp(p-1) - (2p - 2)} + \cdots + (p-1)^{kp(p-1) - (p(p-2) - 2)}\]](http://latex.artofproblemsolving.com/4/7/c/47c88ffecac2c0adb026c6e25e483ab7256f8f34.png)
Note that there are exactly 
terms of each base and they form a geometric progression with common ratio (from right to left).
where 
. Since 
, Fermat's Theorem gives :![\[\frac{x(y^{kp(p-1)} -1)}{y^{p-1}} \equiv 0 \pmod p\]](http://latex.artofproblemsolving.com/8/2/4/824cdd817ccda26478ebd3e843a0847aa53b23a7.png)
If 
, the sum of the geometric progression is just and if 
, the sum is .
.
which gives the desired result.
. We start with the lemma:, we have 
.
where 
is an integer. By FLT we also have 
where is an integer. Then we have:
so 
consists of copies of 
in , implying the desired result. 
. We have:
where we get the second equivalence by double counting, the third by geometric series formula, and the fifth by FLT. such that 
. Then we have:
which is the desired congruence. 
, because the order of any integer mod divides . Then, we can also take all bases mod . Hence, the desired sequence cycles every . groups. Every base is matched to every exponent exactly once -- in particular, if we let be the group, then 
This equals 
because 
for all relatively prime to by Fermat. Hence constructing 
such complete cycles gives us the desired result.
.
.
.
. Since we are adding terms in 
that are all 
, we have 
. Thus, it suffices to show that for all 
such that 
and 
, 
. This is obviously true for 
, so henceforth assume 
., we can rewrite 
as . We see that the exponents in 
cover each distinct residue 
. Since (by FLT) 
, the residue of 
is equal to the residue of 
.
, 
. Since 
is relatively prime to and 
, , as desired.
.
similarly to in Lemma 1, but going all the way up to 
if 
or 
if so that 
.
, the terms we are adding are all , and there are of them. This implies 
. Since it is trivial that 
, it suffices to show that 
, where . cover each residue exactly times. By proceeding similarly to how we did in Lemma 1, we arrive at 
. such that 
and 
, which implies 
.
.. We can see this: 
By FLT we have that: 
if 
.
This implies the claim. 
so 
by the Claim.
and then 
works. 
. Notice that modulo , the bases of the integers in the sum 
must cycle every , though the exponents must cycle every 
since 
for any integer 
relatively prime to . Therefore, we have 
as a "repeating block" for ; that is, 
for any positive integer . Now, we look at 
modulo with the following claim:
. repeat every for each base (where bases are taken modulo ), we can manipulate as follows: 
where 
by Fermat's little theorem. Since 
, this proves the claim. 
be a positive integer such that 
. Combining the claim with our initial observations, we must have 
, so we are done. 
is increased by 
new terms: 
This is what the residue will change by if is increased by . times. Then, since is relatively prime to , each possible residue of 
is the subject of the above expression 
exactly once, since it's what is being increased from. Thus, to find the total change , we sum this over all residues . We will use the fact that the sum of the th powers of the integers mod is 0 unless is divisible by , in which case it is . This means that the sum of when taking mod is just , since everything else vanishes. by will decrease the residue by 1. Thus, by doing so some number of times, we can always reach any residue mod , and we are done.
.
cycle with the base in modulo , we can manipulate it to get
![\[ \sum_{b=1}^{p-1} b^a \equiv
\begin{cases} -1 & p-1 \mid a \\ 0 & p-1 \nmid a \end{cases}
\pmod p \]](http://latex.artofproblemsolving.com/b/7/4/b7421bb5d8181f1e5b53818b062cd173b76e105b.png)
![\[(a+mp)^b \equiv a^b \pmod{p}, \ m \in \mathbb{Z} \ \textbf{ and } \ a^{b+n(p-1)} \equiv a^b \pmod{p}, \ n \in \mathbb{Z} \]](http://latex.artofproblemsolving.com/2/a/2/2a287fe7fed782a1aa9cf1c5d009a4043138fe8e.png)
![\[\implies (a+mp)^{b+n(p-1)} \equiv a^b \pmod{p}, \ m,n \in \mathbb{Z}.\]](http://latex.artofproblemsolving.com/7/7/9/7795d3663567789f60d926c8935552b2fd6b74a7.png)
is equivalent to 
copies of modulo . ![\[f(cp(p-1)) \equiv -c \pmod{p},\]](http://latex.artofproblemsolving.com/d/e/d/ded3176ca7c064c480bf7e4b72681772e28de748.png)
would work. 
is , and the period of the exponent is , the period of the LHS is ..
if 
, and is 
if 
.
![\[1^0 + 2^0 \dots {p-1}^0\]](http://latex.artofproblemsolving.com/6/9/0/6908e2edb311fa98e9ba517613f55b458ccf9a8e.png)
![\[1^1 + 2^1 \dots {p-1}^{1}\]](http://latex.artofproblemsolving.com/a/d/a/ada0160e35efd41ea70470de23dedb5f2942767c.png)
![\[1^2 + 2^2 \dots {p-1}^{2}\]](http://latex.artofproblemsolving.com/1/d/6/1d63fc1393109cae820725d48340499b71c8f000.png)
![\[\dots\]](http://latex.artofproblemsolving.com/1/6/7/16762e47f22f53e2431935f211fa82570b46b6cb.png)
![\[1^{p-1} + 2^{p-1} \dots {p-1}^{p-1}\]](http://latex.artofproblemsolving.com/c/c/8/cc8a2bf65dd0f56300db0244ad309244d4d82e8e.png)
. vanish, and the last row is , so 
. it is obvious that there is some 
so that 
.
terms, the bases cycle, whereas every terms the exponent cycles. Then one ``complete cycle" of the terms modulo occurs every terms. This motivates us to look at one such cycle, and evaluate it modulo . modulo .
Thus consider the partition,
Now modulo this reduces to,
Now note that the last row simply vanishes. The first row simply evaluates as modulo . For all other terms, say with base , we may use the geometric series formula to evaluate the row as 
. This is well defined modulo as the denominator is nonzero ie: 
. Now modulo the numerator is simply ![\[ a\left(a^{p-1} - 1\right) \equiv a\left(1 - 1 \right) \equiv 0 \pmod{p} \]](http://latex.artofproblemsolving.com/8/5/d/85d2b92f40d48b3cb5cfcb7a8ac92f54fe259207.png)
Hence the expression as a whole is simple congruent to 
as claimed. such that our sequence is congruent to 
simply by choosing 
. Thus we cover the whole residue class, and as a result may always find some such that the sum is congruent to 
modulo . is rather arbitrary and as such we should be able to cover the whole residue class. Now to look for some way to control the sequence we look for some such that we can easily evaluate the sequence. Then the easiest such value of is merely when the sequence repeats, or one ``complete cycle".
to be 
. Since is kind of close to the year, we assume it's an arbitrary number chosen by the writers just for funzies. Therefore, instead of only proving that 
such that 
, we claim that this holds true for any integer 
. We will do this by showing the following claim holds true., 
. I will show this in two parts.
.![\[a^k\equiv a^{k+p-1}\mod p.\]](http://latex.artofproblemsolving.com/1/2/6/12619c50e95c4e3ada5b9a645f95ebf1abaccaef.png)
Therefore, note that![\[f(cp(p-1))=1^{cp(p-1)}+\dots+(p(p-1)+1)^{(c-1)p(p-1)}+\dots+(2p(p-1)+1)^{(c-2)p(p-1)}+\dots,\]](http://latex.artofproblemsolving.com/b/b/3/bb3096da8c7f2742f0b2f15ffe0944312f54713a.png)
![\[\equiv (1^{p(p-1)}+\dots+((p(p-1))-1)^1)+(1^{p(p-1)}+\dots+((p(p-1))-1)^1)+\dots,\]](http://latex.artofproblemsolving.com/9/c/8/9c8408e3842781ba7c87f2dd518ce5b429fb0463.png)
![\[\equiv c(1^{p(p-1)}+\dots+((p(p-1))-1)^1), \]](http://latex.artofproblemsolving.com/9/2/0/920872e94a82beb79b4552c7fc064d66ec8e67ed.png)
![\[\equiv cf(p(p-1))\mod p,\]](http://latex.artofproblemsolving.com/2/0/4/20429938f75fec0120b649a581d26902cd66f8d1.png)
which proves the first half of our claim.
.![\[f(p(p-1))=1^{p(p-1)}+\dots+((p(p-1))-1)^1,\]](http://latex.artofproblemsolving.com/5/3/1/531f92f42ae5902b95a5e4712ec901ec4916be19.png)
which we can split into sums![\[1^{p(p-1)}+(p+1)^{p(p-2)}+\dots+(p(p-2)+1)^{p},\]](http://latex.artofproblemsolving.com/9/8/2/982f714b2e4f999551764d28ef5ed1d0b3510b22.png)
![\[\equiv 1+1+\dots \equiv p-1 \equiv -1 \mod p,\]](http://latex.artofproblemsolving.com/d/5/d/d5d67257e09a56ef97bd269dabb9084e06f9cc1d.png)
and for all 
,![\[k^{p(p-1)-k+1}+(p+k)^{p(p-2)-k+1}+\dots+(p(p-2)+k)^{p-k+1},\]](http://latex.artofproblemsolving.com/0/a/6/0a636652eb74dda33c2938e0ed54921e5502ed6e.png)
![\[\equiv (k^{p-k+1})(k^{p(p-2)}+k^{p(p-3)}+\dots+p^0),\]](http://latex.artofproblemsolving.com/8/d/0/8d02a4ff6f52573faa719219e4ea3b0f68ce32b0.png)
![\[=(k^{p-k+1})\left(\frac{k^{p(p-1)}-1}{k^p-1}\right),\]](http://latex.artofproblemsolving.com/1/b/3/1b3c073b2f3a9837b10bde8d2066e4698800a1b0.png)
![\[\equiv (k^{p-k+1})(0) \equiv 0\mod p,\]](http://latex.artofproblemsolving.com/2/7/a/27a9af36eec817069f844a72cb3fabf1676040eb.png)
and for all the multiples of , a mod . Summing these all together, we get a total of![\[f(p(p-1))\equiv p-1+0+0+\dots+0\equiv p-1\mod p,\]](http://latex.artofproblemsolving.com/7/b/5/7b508853ea6f8857f793598dacc9b3623186d57b.png)
meaning that ., finishing the problem. that works, using the fact that , we find that the number 
yields a sum with 
for any integer positive sufficiently large enough so that 
. 
is a valid construction.
I claim that where is a prime.
be a generator of the group 
Also define 
such that 
We can rearrange our sum based on the remainder modulo :
Finally, I claim that 
where is an integer. This is obviously true because every 
the base and the exponent cycle. Thus, for a prime is a valid construction.
,
with 
.
such that 
?
.![\[f(n) = f(p(p - 1)(p - x)) \equiv x \pmod{p} \]](http://latex.artofproblemsolving.com/e/a/2/ea25a4d5886a97cd35c2c0e6307fdde9a846cc56.png)
,
mod 
,
mod 
,
is a complete residue system modulo any prime 
for some positive integer 
,
where the last equality follows from FLT (In particular, 
is congruent to 
for some 
,
modulo 
Since 
works. Then we have:
taking bases ![\begin{align*}
S = k[& 1^0 + 2^0 + \cdots + (p-1)^0 \\ & + 1^1 + 2 ^1 + \cdots + (p-1)^1 \\ & + \cdots \\ & + 1^{p-2} + 2^{p-2} + \cdots + (p-1)^{p-2} ].
\end{align*}](http://latex.artofproblemsolving.com/5/8/9/589e2aef8033fdc775296fa8f07a1cb806fd6b2a.png)
We now prove a lemma
proof. When 
We can replace 
with powers of a primitive root modulo 
Then the sum becomes 
(obviously 
)
then again if we pick a primitve root 
and so our lemma is proved. 
Then if we choose 
we get that 
maneuvering impossible. Say, if 
is written as 
and 
is written as 
,
and 
which is very unpleasant to look at (the variability of primitive root exponents is very unreliable in the local sense.)
where it can be paired up at our will to other classes (or in this case, its own class.) 
.
,![\[ 1^{2p}+2^{2p-1} + 2^{2p-1} + \ldots + p^{p+1} + (p+1)^p + {(p+2)}^{p-1} + \ldots + {(2p)}^1
\]](http://latex.artofproblemsolving.com/6/7/c/67cb297d16dbc566e72a3237266645f613037a84.png)
and at total, this yields 
.
does not repeat itself easily, with the first thing in "base class 1" 
yields 
and the second thing 
yields 
.
every time a new chunk appears, ant that 
of the 
and 
which is 
.
,
![$[0,p-1] \pmod p$](http://latex.artofproblemsolving.com/0/7/1/071b4240253eedc69146a8c97a24763b401f0d41.png)
,
the results follows by Fermat Little Theorem, as every term would be 
contains all residues 
so 
,
and we are done
has period 
,
,![\[m \cdot \sum_{i=1}^{p(p-1)} i^{p(p-1)+1-i} \equiv m \cdot \sum_{i=0}^{p-1} \sum_{j=1}^{p-1} i^j \equiv m(p-1) \pmod p.\]](http://latex.artofproblemsolving.com/d/6/a/d6a93c54ba113bab73e314471f85da9f03f0e278.png)
,
then:
to be 
then we have:
with 
so 
so we get that: