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Source: Mexican Quarantine Mathematical Olympiad P5

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#1 h Apr 26, 2020, 2:03 PM • 3 Y

Y by Math-wiz, Kingsbane2139, AlexCenteno2007

Let $\mathbb{N} = \{1, 2, 3, \dots \}$ be the set of positive integers. Find all functions $f:\mathbb{N}\rightarrow \mathbb{N}$ , such that for all positive integers $n$ and prime numbers $p$ :
$p \mid f(n)f(p-1)!+n^{f(p)}.$
Proposed by Dorlir Ahmeti

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#2 h Apr 26, 2020, 3:35 PM • 4 Y

Y by Math-wiz, amar_04, Pluto1708, Isluna

Nice and easy! Here's my solution: The only such function is the identity function, which can be easily seen to work using Wilson's Theorem and FLT. Now we show that this is the only solution. Suppose $f(1)>1$ . Then there exists a prime $p$ dividing $f(1)$ . But this gives $0 \equiv f(1) \cdot f(p-1)! \equiv -1 \pmod{p}$ which is impossible. So we have $f(1)=1$ . Then for $n=1$ , we get $f(p-1)! \equiv -1 \pmod{p}$ Thus, we can rewrite the given condition as $P(n,p) := f(n) \equiv n^{f(p)} \pmod{p} \quad \forall n \in \mathbb{N},p \in \mathbb{P}$ Now, if some prime $q \neq p$ divides $f(p)$ , then we have $P(p,q) \Rightarrow 0 \equiv f(p) \equiv p^{f(q)} \pmod{q}$ which cannot happen for distinct primes $p,q$ . This means that $f(p)$ is a power of $p$ for all primes. But this gives $f(p) \equiv 1 \pmod{p-1} \Rightarrow n^{f(p)} \equiv n \pmod{p}$ Using $P(n,p)$ we get that $p \mid f(n)-n$ for all $p \in \mathbb{P}$ . This can only happen if $f(n)=n$ , as desired. $\blacksquare$

This post has been edited 2 times. Last edited by math_pi_rate, Apr 26, 2020, 3:40 PM
Reason: $\Rightarrow$ instead of $\rightarrow$

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Functional_equation

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Functional_equation

#3 h Apr 26, 2020, 4:20 PM • 1 Y

Y by AlexCenteno2007

Woow,nice problem!
Claim 1:
$p-1\geq f(p-1)$

Proof:
$P(1,p)\implies p|f(1)f(p-1)!+1\implies p\nmid f(p-1)!$
$Then$
$p-1\geq f(p-1)$

Claim 2:
$p|f(pk)$

Proof:
$P(pk,p)\implies p|f(pk)f(p-1)!+(pk)^{f(p)}$
$Then$
$p|f(pk)$

Claim 3:
$n-odd\implies f(n)-odd$
$n-even\implies f(n)-even$

Proof:
$p-1\geq f(p-1)$
$p=2\implies 1\geq f(1)\implies f(1)=1$
$p=3\implies 2\geq f(2),2|f(2)\implies f(2)=2$

$P(n,2)\implies 2|f(n)f(1)!+n^{f(2)}\implies 2|n^2+f(n)$
$Then$
$n-odd\implies f(n)-odd$
$n-even\implies f(n)-even$

Claim 4:
$f(p-1)\geq p-1$

Proof:
$p>2\implies f(p)-odd$
$P(p-1,p)\implies p|f(p-1)f(p-1)!+(p-1)^{f(p)}$
$f(p)-odd\implies p|f(p-1)f(p-1)!-1$
$P(1,p)\implies p|f(p-1)!+1$
$Then$
$p|f(p-1)f(p-1)!-1+f(p-1)!+1\implies p|f(p-1)+1$
$Then$
$f(p-1)+1\geq p\implies f(p-1)\geq p-1$

Claim 5:
$f(p-1)=p-1$

Proof:
$p-1\geq f(p-1),f(p-1)\geq p-1\implies f(p-1)=p-1$

Claim 6:
$p-1|f(p)-1$ $or$ $f(p)-1|p-1$

Proof:
$P(n,p)\implies p|f(n)(p-1)!+n^{f(p)}$
$And$
$(p-1)!\equiv -1 (mod p)$
$Then$
$p|n^{f(p)}-f(n)$
$P(2,p)\implies p|f(2)f(p-1)!+2^{f(p)}$
$Then$
$p|2^{f(p)-1}-1$
$Then$ (Order and Fermat theorem)
$p-1|f(p)-1$ $or$ $f(p)-1|p-1$

Case 1:
$f(p)-1|p-1\implies p\geq f(p)$
$Claim2\implies f(p)\geq p$
$Then$
$f(p)=p$
$Then$
$p|n^{p}-f(n)\implies p|n-f(n)$
$p-biggest\implies n=f(n)$

Case 2:
$p-1|f(p)-1\implies f(p)\equiv 1(mod p)$
$Then\implies Fermat$
$n^{f(p)}-f(n)\equiv n-f(n)(mod p)$
$p-biggest\implies n=f(n)$

This post has been edited 2 times. Last edited by Functional_equation, Apr 26, 2020, 4:21 PM

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#4 h Apr 26, 2020, 4:47 PM • 3 Y

Y by Mathological03, GorgonMathDota, Sugiyem

Good practice on Function Divisibility!
The only function is $f(n) = n$ , which truly satisfied by Wilson and Fermat Little Theorem. We'll prove that this is the only function.
Let $P(n,p)$ be the assertion of $n,p$ to the given functional equation.
Claim 01. $f(p - 1) \le p - 1$ .
Proof. Suppose otherwise, then $f(p - 1) \ge p$ , forcing $p | n^{f(p)}$ for all $n$ , which is not true.

Claim 02. $f(p)$ is a power of $p$ .
Proof. $P(1,p)$ gives us $f(p - 1)! \equiv -1 \ (\text{mod} \ p)$ . Therefore, we have the equivalent assertion of
$n^{f(p)} \equiv f(n) \ (\text{mod} \ p)$ Now, let $p$ be a prime divisor of $n$ , then we have $p | f(n)$ , or generally $\text{rad}(n) | f(n)$ for all $n$ .
Now, suppose that there exists another prime number $q \not= p$ dividing $f(p)$ . Therefore,
$0 \equiv f(p) \equiv p^{f(q)} \ (\text{mod} \ q)$ forcing $q | p$ , or $q = p$ .
Therefore, $f(p)$ is a power of $p$ .

Let $f(p) = p^k$ . Therefore, since $p - 1 | f(p) - 1$ , which means
$p | n - f(n)$ for all $n,p$ . Take $p$ sufficiently large forces $f(n) = n$ for all $n \in \mathbb{N}$ .

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#5 h Apr 26, 2020, 6:06 PM

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Isn't Dangerousliri from Albania? How come could he propose to a Mexican contest.

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#6 h Apr 26, 2020, 6:47 PM • 2 Y

Y by plagueis, Alireza_Amiri

Mathological03 wrote:

Isn't Dangerousliri from Albania? How come could he propose to a Mexican contest.

Actually I'm from Kosovo and we are close with Albania. One friend from Mexico did ask me if I have a problem for the contest, since they were missing some good number theory, so I did create this problem especially for him. Also the geometry problem they were missing some easy problem. They did have but even there were to easy or a bit hard for problem 1. So I did remember that I had one problem that I did save it so then I did propose it and it was perfect for problem 1.

This post has been edited 1 time. Last edited by dangerousliri, Apr 26, 2020, 6:49 PM

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#8 h Apr 26, 2020, 7:41 PM

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Mathological03 wrote:

Isn't Dangerousliri from Albania? How come could he propose to a Mexican contest.

Maybe he's Mexican in secret.

Let $P (n)$ be the assertion $p\mid f (n)f (p-1)!+n^{f (p)}$ .

$P (n)\implies f (n)f (p-1)!\equiv -n^{f (p)}\mod p$ so it follows the set of primes of $f (n)$ are the same of $n$ i.e since $1$ has no primes divisors we must have $f (1)=1$ so $f (p-1)!\equiv -1\mod p$ .

$P (p-1)\implies f (p-1)\equiv (-1)^{f(p)}$ but since $f (p)=p^k$ we must have $p\mid f (p-1)+1$ for any prime. But since $f (p-1)<p$ it follows $f (p-1)=p-1$ .

(We can use directly $FLT$ but this approach is achieved above)

Assume $f (x_1)=f (x_2)$ for some $x_1\neq x_2$ so
$P (x_1)~\text {and}~P (x_2)\implies f (x_1)\equiv x_1^{f (p)}\equiv x_2^{f(p)}\mod p$ so $x_1=x_2$ contradiction.

Now assume assume for a $k$ it follows $f (k)>f(m):\forall~1\le m<k$ . So $P (k+1)\implies f (k+1)-f (m)\equiv (k+1)^{f (p)}-m^{f (p)}\mod p$ . So $f (k+1)>f (k)$ .

For any large $p$ we always we have a bounded, so from $f (p-1)>\cdots>f (1)$ we always we obtain $f (n)=n$ .

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#9 h Apr 26, 2020, 8:21 PM

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Al3jandro0000 wrote:

Mathological03 wrote:

Isn't Dangerousliri from Albania? How come could he propose to a Mexican contest.

Maybe he's Mexican in secret.

No I'm not, I explained above.

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1218 posts

#10 h Apr 26, 2020, 8:47 PM • 1 Y

Y by amar_04

Al3jandro0000 wrote:

Let $P (n)$ be the assertion $p\mid f (n)f (p-1)!+n^{f (p)}$ .

$P (n)\implies f (n)f (p-1)!\equiv -n^{f (p)}\mod p$ so it follows the set of primes of $f (n)$ are the same of $n$

That's not true. In particular, it might be possible that some prime divisor of $n$ divides $f(p-1)!$ instead of $f(n)$ . You will have to first show that $f(p-1)! \equiv -1 \pmod{p}$ , and then conclude this.

Functional_equation wrote:

Claim 6:
$p-1|f(p)-1$ $or$ $f(p)-1|p-1$

Proof:
$P(n,p)\implies p|f(n)(p-1)!+n^{f(p)}$
$And$
$(p-1)!\equiv -1 (mod p)$
$Then$
$p|n^{f(p)}-f(n)$
$P(2,p)\implies p|f(2)f(p-1)!+2^{f(p)}$
$Then$
$p|2^{f(p)-1}-1$
$Then$ (Order and Fermat theorem)
$p-1|f(p)-1$ $or$ $f(p)-1|p-1$

The last line is untrue. For eg. $2^9 \equiv 1 \pmod{7}$ , but $9 \nmid 6$ and $6 \nmid 9$ .

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#11 h May 7, 2020, 12:16 PM

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isn't this too easy for P5 :blush:

:blush:

let $P(n,p)$ the assertion $p \mid f(n)f(p-1)!+n^{f(p)}.$

claim(1): $f(1)=1$
proof:
suppose $f(1)>1 \implies 2|f(1)!$ but $P(1,2) \implies 2|f(1)f(1)!+1 \implies f(1)!$ is odd
contradiction $\blacksquare$
$P(1,p) \implies p|f(p-1)!+1$
now if $p|f(n)$ we have $p|n$
so $f(p)=p^a$
but we have $n^{f(p)} \equiv n^{p^a}\equiv n \bmod p$
so $p|f(n)-n \forall n \in \mathbb{N}$ take $p \rightarrow \infty \implies$ $f(n)=n \forall n \in \mathbb{N}$ and we win

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#12 h Aug 10, 2020, 7:47 PM

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Just another way to end it. We use Dirichlet theorem. Let $g$ be a primitive root modulo $p$ . After we prove that the original equation is equivalent to $p | n^{f(p)} - f(n)$ . We know that $f(p-1) = p-1$ after what @above did. Now by dirichlet we know that exists a prime number such that $g + 1 + kp = q$

By the divisibility $p | (g + kp)^{f(p)} - (g + kp)$ , and we have that $g^{f(p)} \equiv g$ modulo $p$ . As $p| f(p)$ , we have that $f(p) =p$ .

Now $p | n - f(n)$ for all $p$ prime, and now $f(n) = n$ , as desired.

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#13 h Mar 25, 2021, 11:46 PM

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The only answer is $f(n)=n$ , for every natural $n$ .

First we notice that if $f(p-1) \geq p$ , then we must have that $n^{f(p)} \equiv 0 \; \pmod{p}$ , for every $n$ , but this isn't obviously true for every $n$ .
Thus we must have that $f(p-1) < p$ .
Thus we conclude that $f(1)=1$ .

Now we must have that:
$p \mid f(p-1)!+1$ thus by Wilson's theorem we must have that $f(p-1)=p-1$ .

By setting $n=p$ , we must have that $p \mid f(p)$ .

Now choose $n=q-1$ , where $q$ is a prime number and $(q-1,p)=1$ .
This clearly gives us:
$(q-1)^{f(p)-1} \equiv 1 \pmod{p}$ this implies that $p-1 \mid f(p)-1$ .

Combining this with $p\mid f(p)$ , we must have that $f(p)-p=g(p)p(p-1)$ , where $g$ is a function from the naturals to itself.
This gives us that $f(p)=p(t(p)(p-1)+1)$ .

Now set $n=q$ , where $q$ is a prime number different from $p$ and $(q-1,p)=1$ .
Then we easily get the relation:
$(p-1)!t(q)q(q-1) \equiv 0 \pmod{p}$ Since we have a lot of choices for $p$ , this gives us that $t(q)=0$ .
Thus we have that $f(p)=p$ .

With that in mind we have that:
$f(n) \equiv n \pmod{p}$ Now just choose a prime number $p$ s.t. $p > f(n)$ , thus giving us $f(n)=n$ , for every natural $n$ .

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#14 h Oct 25, 2021, 8:42 AM

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plagueis wrote:

Let $\mathbb{N} = \{1, 2, 3, \dots \}$ be the set of positive integers. Find all functions $f:\mathbb{N}\rightarrow \mathbb{N}$ , such that for all positive integers $n$ and prime numbers $p$ :
$p \mid f(n)f(p-1)!+n^{f(p)}.$
Proposed by Dorlir Ahmeti

Cute, but very easy.

Let $P(p,n)$ denote the assertion of the functional equation. $P(2,1)$ gives us that $f(1)f(1)!$ is odd, so $f(1)=1$ . $P(p,1)$ then gives us that $(p-1)!+1\mid f(p-1)!+1$ and so $f(p-1)!\geq(p-1)!$ . Assume that $f(p-1)>p-1$ . Then, $p\mid f(p-1)!$ , so the original functional equation becomes $p\mid n^{f(p)}$ Choosing $n$ coprime to $p$ gives a contradiction. Therefore $f(p-1)=p-1$ for all primes $p$ . Therefore, the divisibility condition gives us that $p\mid n^{f(p)}-f(n)$ due to Wilson. Therefore, the divisibility is reduced to $p\mid n-f(n)$ for all primes $p$ and all positive integers $n$ . Fixing $n$ and varying $p$ gives us that $f(n)=n$ for all $n\in\mathbb{N}$ .

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#15 h Oct 25, 2021, 3:50 PM

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Quite dangerous for me!
Here $p$ is a prime throughout the solution.
Let $d(p,n)$ be the given assertion.
Claim: $f(1)=1$
Proof: Assume $f(1) \ge 2$ . So there exists a prime $p_0$ s.t. $p_0 \mid f(1)$ .
Then $d(p_0,1) \implies p_0 \mid f(1)f(p-1)!+1$
$\implies p_0 \mid 1$ which is a contradiction. Hence $f(1)=1$
Thus we get $p \mid f(p-1)!+1$
So new $d(p,n)$ is $p \mid n^{f(p)}-f(n)$
Claim: $p \mid f(p)$
Proof: $d(p,p) \implies p \mid p^{f(p)}-f(p) \implies p \mid f(p)$
Claim: $f(p)=p^m$ where $m\in\mathbb{N}$
Proof: Assume FTSOC that there exists a prime $q$ (different from $p$ ) such that $q \mid f(p)$ .
Then $d(q,p) \implies q \mid p^{f(p)}-f(p) \implies q \mid p$ which is a contradiction. So by the fundamental theorem of numbers, $f(p)=p^m$ .
Hence, $n^{f(p)} \equiv n^{f(p) (\mod (p-1))} \equiv n^1 (\mod p)$ $\implies p \mid n^{f(p)}-n$ So $d(p,n) \implies p \mid f(n)-n$
So the number $f(n)-n$ has infinitely many divisors which implies $f(n)-n=0 \implies \boxed{f(n)=n}$ is the only solution. $\blacksquare$

This post has been edited 2 times. Last edited by sanyalarnab, Oct 25, 2021, 3:51 PM
Reason: :)

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#16 h Dec 27, 2021, 12:08 PM

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Primitive root of $p$ and Drichlet theorem

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F10tothepowerof34

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F10tothepowerof34

#18 h Jul 14, 2023, 12:44 PM

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Let $P(n,p)$ denote $p\mid f(n)f(p-1)!+n^{f(p)}$

$P(1,2)$ yields $2\mid f(1)f(1)!+1\Longleftrightarrow f(1)f(1)!\equiv 1\pmod 2$ , however notice that if $f(1)\ge2$ , we obtain $f(1)!\equiv 0\pmod 2$ which is a contradiction.
Thus $f(1)=1$

$P(1,p)$ yields $p\mid f(1)f(p-1)!+1\Longrightarrow p\mid f(p-1)!+1\Longleftrightarrow f(p-1)!\equiv -1\pmod p$
Therefore the original assertion transforms into $p\mid n^{f(p)}-f(n):=Q(n,p)$

Furthermore $P(p,p)$ yields $p\mid f(p)f(p-1)!+p^{f(p)}$ moreover since $p\mid p^{f(p)}$ , $p$ must also divide $f(p)f(p-1)!, p\mid f(p)f(p-1)!\Longleftrightarrow f(p)f(p-1)!\equiv 0\pmod p$ , however notice that $f(p)f(p-1)!\equiv -f(p)\pmod p$ thus $f(p)\equiv 0\pmod p\Longrightarrow p\mid f(p)$

Now, $FTSOC$ assume $\exists q\text{ such that it is a prime and }q\neq p\text{ and }q\mid f(p)$
Therefore from $P(p,q)$ we obtain $q\mid f(p)f(q-1)!+p^{f(q)}\Longrightarrow q\mid p^{f(q)}\Longrightarrow q\mid p$ which is clearly a contradiction. Thus $f(p)$ doesn't have distinct prime factors, therefore $f(p)=p^a$ for some $a\in\mathbb{Z}^+$

Now the assertion $Q(n,p)$ transforms into $p\mid n^{p^a}-f(n)$ , however notice that $n^{p^a}\equiv n\pmod p$ thus $n^{p^a}-f(n)\equiv n-f(n)\equiv 0\pmod p$ which forces $f(n)=n, \forall n\in\mathbb{N}$

In conclusion $\boxed{f(n)=n, \forall n\in\mathbb{N}}$ $\blacksquare$ .

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#19 h Mar 31, 2025, 7:40 AM

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Looked like a monster, but was certainly easy

Here's my solution.

Attachments:

Mexico Quarantine Math Oly P5.pdf (29kb)

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cursed_tangent1434

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cursed_tangent1434

#20 h Mar 31, 2025, 11:36 AM

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Pretty straightforward and standard but still not easy per se. The answer is $f(n) = n$ for all $n\in \mathbb{N}$ . It’s easy to see that these functions satisfy the given equation. We now show these are the only solutions. Let $P(n,p)$ denote the assertion that $p \mid f(n)f(p-1)!+n^{f(p)}$ for a prime $p$ and natural number $n$ .

First note that $P(1,2)$ implies $2\mid f(1)f(1)!+1$ which is impossible if $f(1)\ge 2$ so we must have $f(1)=1$ .

Claim : We have $f(p-1)=p-1$ for all odd primes $p$ .

Proof : Consider an odd prime $p$ . Note $P(1,p)$ implies
$p \mid f(1)f(p-1)!+1^{f(p)}=f(p-1)!+1$ Thus, $f(p-1)! \equiv -1 \pmod{p}$ and in particular we must have $f(p-1)<p$ . Next, $P(p-1,p)$ yeilds
$p\mid f(p-1)f(p-1)!+(p-1)^{f(p)} \equiv -f(p-1)+(-1)^{f(p)}$ which implies that $f(p-1) \equiv 1 \pmod{p}$ or $f(p-1) \equiv -1 \pmod{p}$ . However, since $f(p-1)<p$ this implies $f(p-1) \in \{1,p-1\}$ of which the first is impossible since $f(p-1)! \equiv -1 \pmod{p}$ . Thus, $f(p-1)=p-1$ for all primes $p$ as desired.

Claim : We have $f(p)=k_p(p-1)+1$ for some positive integer $k_p$ for each odd prime $p$ .

Proof : Let $g$ be a primitive root $\pmod{p}$ . By Dirichlet's Theorem we have that there exists some prime $q$ such that $q-1 \equiv g \pmod{p}$ . But then, $P(q-1,p)$ implies
$p\mid f(q-1)f(p-1)!+(q-1)^{f(p)} \equiv (q-1)^{f(p)}-(q-1) \equiv g(g^{f(p)-1}-1)$ This implies that $p-1 \mid f(p)-1$ which implies that $f(p)=k_p(p-1)+1$ for some positive integer $k_p$ as claimed.

Now,
$p\mid f(n)f(p-1)! + n^{f(p)} \equiv n^{k_p(p-1)+1}-f(n) \equiv n-f(n)$ which implies that $p \mid n-f(n)$ for all odd primes $p$ . But, this immediately implies that $f(n)-n=0$ and thus $f(n)=n$ for all positive integers $n$ as desired.

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alexanderhamilton124

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alexanderhamilton124

#21 h Apr 1, 2025, 11:32 AM • 1 Y

Y by L13832

Note that if $f(p - 1) \geq p$ for any $p$ , consider $p \nmid n$ , and it fails. Hence, $f(p - 1) < p \implies f(1) < 2 \implies f(1) = 1$ . Taking $n = 1$ , we have $f(p - 1)! \equiv -1\mod{p} \implies p \mid n^{f(p)} - f(n)$ .

This means that $q \mid f(n) \Leftrightarrow q \mid n$ . Consider $n = r$ , where $r$ is a prime. This means that $f(r)$ is a power of $r$ , for all primes. Consider $r > f(n) + n$ for any positive integer $n$ . We have $r \mid n^{f(r)} - f(n)$ . Note that $n^{f(r)} \equiv n \mod{r} \implies r \mid n - f(n) \implies f(n) = n$ . We are done.

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