Let $x,y,z$ be pairwise coprime positive integers and $p\ge5$ and $q$ be prime numbers which satisfy the following conditions: (i) $6p$ does not divide $q-1$; (ii) $q$ divides $x^2+xy+y^2$; (iii) $q$ does not divide $x+y-z$. Prove that $x^p+y^p\ne z^p$.
Problem
Source: Bulgaria 1989 P6
Tags: number theory, Divisibility
Mathological03
14.06.2021 20:18
Overkill : FLT for the win
jasperE3
14.06.2021 20:20
Note that this competition was held before Andrew Wiles proved FLT.
Sprites
15.06.2021 08:11
tough work without flt
Let's prove it by contradiction,
If x^p+y^p=z^p
Now we divide into casework.
a)p is even:
If p was even then p∤q and also x²ⁿ+y²ⁿ=z²ⁿ.
Now since q∤x+y-z we can see that either x+y is not congruent to z(mod q)
So x+y is not congruent to z mod q² either so taking mod q² to x²ⁿ+y²ⁿ=z²ⁿ we get x+y congruent to z mod q² which is a contradiction.hence no solutions exist.
b)p is odd
If p was odd x^p+y^p=z^p then (x+y)|z which means that q∤x+y .
Now use fermat's last theorem , x^(q-1)+y^(q-1)≡1(mod q )
now this means z^p≡2(mod q)
Now since 6p∤q-1 or p∤q-1
Which provides a contradiction mod q .
Hence no solutions exist.
ihategeo_1969
24.08.2024 04:27
Thank you Andrew Wiles.
grupyorum
13.10.2024 16:20
Let's crack this. First, from the fact $q\mid x^2+xy+y^2$, we deduce $q\mid (2x+y)^2 + 3y^2$, i.e., $(-3/q)=1$. So, $q=3$ or $q\equiv 1\pmod{6}$ (a well-known fact). Let $q=3$. Then $q\nmid x,y$ but $3\mid x^2+xy+y^2$, so $x\equiv y\pmod{3}$. Moreover, $z\not\equiv 2x\pmod{3}$. Now if $x^p+y^p=z^p$, then $x^p+y^p\equiv 2x^p \equiv (2x)^p\pmod{3}$, yielding $3\mid 2x-z$, a contradiction. So, no solutions if $q=3$.
Write $(x+y)^p = (x+y)(x^2+2xy+y^2)^{(p-1)/2}\equiv (x+y)(xy)^{(p-1)/2}\pmod{q}$, set $p=6\ell\pm 1$ and do some casework.
(based on the values of $p$ modulo 6) reveals $(x+y)^p\equiv x^p+y^p\pmod{q}$. So, if $x^p+y^p=z^p$, then $(x+y)^p\equiv z^p\pmod{q}$. Check that if $q\mid z$, then $q\mid x+y$, contradicting with $q\nmid x+y-z$. So, $q\nmid x+y$, $q\nmid z$, and therefore, $(x+y)^{q-1}\equiv z^{q-1}\pmod{q}$ by Fermat's (not the last, but the little) Theorem. Hence, we arrive at $(x+y)^d\equiv z^d\pmod{q}$, where $d={\rm gcd}(p,q-1)\in\{1,p\}$. Since $p\nmid q-1$, we have $d=1$, so $q\mid x+y-z$, a contradiction. Hence, no solutions.