Pythagorean-Euler Theorem?

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Source: 2022 IMOC N5

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#1 h Sep 5, 2022, 1:38 PM • 1 Y

Y by Mango247

Find all solution $(p,r)$ of the "Pythagorean-Euler Theorem" $p^p+(p+1)^p+\cdots+(p+r)^p=(p+r+1)^p$ Where $p$ is a prime and $r$ is a positive integer.

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This post has been edited 2 times. Last edited by CrazyInMath, Sep 9, 2022, 12:53 PM

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#2 h Sep 8, 2022, 10:12 AM • 4 Y

Y by GuvercinciHoca, Mango247, Mango247, Mango247

Answer: $(p,q)=(3,2)$ is the only solution.+
First consider the case where $p=2$ .
The equation becomes $2^2+..+(2+r)^2=(3+r)^2\Leftrightarrow \dfrac{(2+r)(3+r)(2r+5)}{6}-1=(r+3)^2$
so $r+3|6\Rightarrow r=3$ , but $2^2+3^2+4^2+5^2\neq 6^2$ so no solutions.
Therefore $p\geq 3$
Let $q$ be an odd prime divisor of $r+1$ . Then $r+1=lq$ for some positive integer $l$
Since $p,p+1,..,p+r$ are $r+1$ consecutive positive integers we deduce that
$p^p+(p+1)^p+..+(p+r)^p\equiv l(1^p+2^p+..+(q-1)^p) \pmod q$
Now let $g$ be a primitive root modulo $q$ . Then $1^p+..+(q-1)^p\equiv g^p+(g^p)^2+..(g^p)^{q-1} \pmod q$
Observe that if $g^p\equiv 1 \pmod q$ then we must have $q-1|p$ which is impossible since $p$ is odd.
Hence $g^p\not \equiv 1 \pmod q$ and $g^p+(g^p)^2+..(g^p)^{q-1}\equiv \dfrac{g^p((g^p)^{q-1}-1)}{g^p-1}\equiv 0 \pmod q$
So $q|p^p+(p+1)^p+..+(p+r)^p=(p+r+1)^p\equiv p^p \pmod q$ so $q=p$ .
As a result $r+1=2^bp^a$ for some $a,b\geq 0$
If $4|r+1$ then
$1\equiv (p+r+1)^p=p^p+(p+1)^p+..+(p+r)^p\equiv p+(p+1)+..+(p+r) \equiv (r+1)p+\dfrac{r(r+1)}{2}\equiv 0\pmod 2$ a contradiction.
So $r+1=p^a$ or $r+1=2p^a$
Suppose that $r+1=2p^a,a\geq 0$
If $a=0$ then $p^p+(p+1)^p=(p+2)^p\Rightarrow 1=2^p\pmod p$ a contradiction.
So $a\geq 1$
Let $q$ be a prime divisor of $r$ , $q\neq 2$ since $r+1$ is even. Let $r=lq$
$p^p+(p+1)^p+..+(p+r)^p\equiv l(1^p+..+(q-1)^p)+p^p \pmod q$ , as before $q|1^p+..+(q-1)^p$
so $p^p\equiv (p+1)^p \pmod q\Leftrightarrow q|z^p-1$ where $z\equiv p(p+1)^{-1} \pmod q$
Now $q|z^p-1 \Leftrightarrow ord_qz|p$ so $ord_qz=1$ or $ord_qz=p$
If $ord_qz=1$ we get that $p=p+1\pmod q$ which is impossible.
So $p= ord_qz|q-1$ . Hence if we write $r=2p^a-1=q_1^{a_1}q_2^{a_2}..q_2^{a_t}$ we have that $q_i|r$ and thus $q_i=1\pmod p$ for $i=1,2,..,t$
This implies that $-1=2p^a-1=1\cdot1..\cdot 1=1 \pmod p$ which means $p|2$ a contradiction.

It remains the case where $r+1=p^a, a\geq 1$
Using the same argument as before (take an odd prime factor $q$ of $r$ ) we get that $r=p^a-1=2^mq_1^{a_1}q_2^{a_2}..q_2^{a_t}\equiv 2^m\pmod p$
so $p|2^m+1$ and $2^m|p^a-1$
Now take an odd prime factor $q$ of $r+2$ . Let $r+2=ql$
Then $(p-1)^p\equiv (p+r+1)^p=p^p+(p+1)^p+..+(p+r)^p\equiv l(1^p+..+(q-1)^p)-(p+r+1)^p\equiv-(p-1)^p \pmod q$
so $q|p-1$ , but also $q|r+2=p^a+1$ thus $q|2$ a contradiction.
As a result $r+2=2^t$ for some $t\geq 2$ .
$2^m|p^a-1=2^t-2\Rightarrow m=1$ so $p|2^m+1=3\Rightarrow p=3$ and now $3^a+1=2^t$
Since $4|3^a+1$ we get that $a$ is odd and thus $3^a+1=4\pmod 8$ so $t=2$ and $r=2$
Indeed $3^3+4^3+5^3=6^3$ so $(p,r)=(3,2)$ is the only solution of the ”Pythagorean-Euler Theorem”

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