Find all prime numbers $p$ for which there exist positive integers $x$, $y$, and $z$ such that the number $x^p + y^p + z^p - x - y - z$ is a product of exactly three distinct prime numbers.
Problem
Source: 2019 Junior Balkan MO
Tags: number theory, Junior, JBMO, Balkan, 2019, prime numbers
22.06.2019 15:20
I did try a little bit this problem and I did solve it but I'm lazy to write all maybe there is a shorter way to do it. First for $p=2$ take $(x,y,z)=(1,1,7)$, for $p=3$ take $(x,y,z)=(1,2,3)$ and for $p=5$ take $(x,y,z)=(1,1,2)$. Now for $p\geq 7$ I did prove that $x^p+y^p+z^p-x-y-z$ is divisible with $2,3,p$ hence it should be equal with $6p$. So $x^p+y^p+z^p-x-y-z=6p$ from here I bashed with $AM-GM$ to try that this is not possible.
22.06.2019 16:41
I will continue the thought of @dangerousliri. We will prove that $a^p-a\geq 2p$, with $a>1$. We can just do induction. For $p=7$ the statement holds. Let it is true for $p=q$. We will prove that it holds for $p=q+1$. $a^{p+1}-x>2(p+1)$. But we know that $a^p-a>2p$. So, we want to prove that $a^{p+1}-a>a^p-a+2\Leftrightarrow a^p(a-1)>2$, which is true for $a>2$
22.06.2019 16:48
minageus wrote: I will continue the thought of @dangerousliri. We will prove that $a^p-a\geq 2p$, with $a>1$. We can just do induction. For $p=7$ the statement holds. Let it is true for $p=q$. We will prove that it holds for $p=q+1$. $a^{p+1}-x>2(p+1)$. But we know that $a^p-a>2p$. So, we want to prove that $a^{p+1}-a>a^p-a+2\Leftrightarrow a^p(a-1)>2$, which is true for $a>2$ You should also try what happens when $a=1$. I think more good way would be this way. Since $x,y,z$ can't be all equal with $1$ we can suppose $x\geq 2$ and since $y^p\geq y$ and $z^p\geq z$ we left to show $x^p-x>6p$ which is true (try it to show this by AM-GM and then induction) and deliver contradiction.
22.06.2019 16:56
Yes, ok, we see that we cannot have all the numbers equal to 1, so maybe we shoud use induction in the initial inequality, without breaking it to parts.
22.06.2019 17:25
Also, we must say that we can show that $x^p+y^p+z^p-x-y-z$ is divisible by $2,3,p$ by using Fermat's Little Theorem.
22.06.2019 18:36
This problem was proposed by (Silouanos Brazitikos) me
22.06.2019 20:44
silouan wrote: This problem was proposed by (Silouanos Brazitikos) me Congratulations, Mr Brazitikos. I hope that the Greek team will have the desired results under your leadership!
23.06.2019 08:25
silouan wrote: This problem was proposed by (Silouanos Brazitikos) me Very nice problem, although it might be a bit difficult for Problem No 1. Congratulations
23.06.2019 14:34
delegat wrote: silouan wrote: This problem was proposed by (Silouanos Brazitikos) me Very nice problem, although it might be a bit difficult for Problem No 1. Congratulations A bit difficult???? This belongs in the AMC 8 Shortlist
23.06.2019 20:15
frog_man wrote: A bit difficult???? This belongs in the AMC 8 Shortlist Well, from 57 students of the official countries, only 20 of them got a 10/10, so for Problem 1 it turned out to be a bit difficult.
11.01.2020 04:16
The answer is $p=2, 3, 5$. We can take $(x, y, z) = (1, 1, 6), (1, 2, 3), (1, 1, 2)$ respectively to show that these work. Now suppose $p>5$. By FLT we have $x^p -x \equiv 0 \pmod{p}, x^p-x \equiv 0 \pmod{2}, x^p-x \equiv 0 \pmod{3}$, so we have $x^p + y^p + z^p - x - y - z \equiv 0 \pmod{6p}$. But since at least one of $x, y, z$ must be greater than $1$, we have $x^p + y^p + z^p - x - y - z \geq 2^p-2 > 6p$ if $p \geq 7$; thus since $6p$ is a proper divisor of $x^p + y^p + z^p - x - y - z$, we are done.
30.06.2022 13:37
The answer is $\boxed{2,3,5}$. Proof of sufficiency: For $p=2$, $p=3$, and $p=5$, choose $(x,y,z) = (1,1,6), (1,2,3), (1,1,2)$ respectively, which clearly work. Proof of necessity: Suppose $p>5$. Note that $x^p\equiv x\pmod p, y^p\equiv y\pmod p, z^p\equiv z\pmod p$, so $p\mid x^p+y^p+z^p-x-y-z$. Claim: For any odd prime $p$ and positive integer $a$, we have $a^p\equiv a\pmod 2$ and $a^p\equiv a\pmod 3$. Proof: If $2\mid a$, then $a^p $ and $a$ are both even. If $2\nmid a$, then $a^p$ and $a$ are both odd. So $a^p\equiv a\pmod 2$. If $3\mid a$, then $3\mid a^p$. If $3\nmid a$, then $a^{p-1}\equiv 1\pmod 3$, so $a^p\equiv a\pmod 3$. $\square$ Thus, $6\mid x^p+y^p+z^p-x-y-z$, which implies $x^p+y^p+z^p-x-y-z=6p$. Since at least one of $x,y,z$ is greater than $1$, we have \[x^p+y^p+z^p-x-y-z=(x^p-x)+(y^p-y) + (z^p-z) \ge 2^p-2 > 6p\]for $p>7$, contradiction.
10.09.2022 19:45
The answers are $p=2,3,5$. Choose $(x,y,z)=(2,7,11),(1,2,3),(2,1,1)$ for $p=2,3,5$, respectively. Now, we assume the contrary that $p$ can be at least $7$. Note that $2,3,p\mid x^p-x, y^p-y, z^p-z$, and so $$(x^p-x)+(y^p-y)+(z^p-z)=6p.$$Note that $\frac{d}{dx} (x^p-x)=px^{p-1}-1>0$, so it is increasing. But by induction, we have $2^p-2>6p$ for all $p\geq 7$. $$(x^p-x)+(y^p-y)+(z^p-z)>(2^p-2)+0+0>6p$$This gives us a contradiction.