No solutions of $x^n+y^n=z^n$ such that $x$, $y$, $z$ are primes

Question

Problem. Show that for $n\ge 2$ there are no solution $$x^n+y^n=z^n$$ such that $x$, $y$, $z$ are prime numbers.

Personally I'd consider this a relatively cute problem which can be given to students when talking about Fermat's Last Theorem - and which should be relatively easily solvable. (I can post my solution - but I suppose that the solutions which will be given here are very likely to be cleverer than mine.)

I will stress that we're looking that the solutions where simultaneously all three numbers are primes - unlike a more difficult problem posted here: Diophantine Equation $x^n + y^n =z^n (x<y, n>2)$.

I have searched on the site a bit to see whether this problem has been posted here before. I only found this deleted question: How we can deal with this equation $a^n+b^n=c^n$ if it was given to have solutions in primes numbers not integers numbers? (Of course, it is quite possible that I might have missed something. After all, searching on this site is not easy.)

Answer 1

Clearly, $x=y$ is impossible as $x\mid 2x^n=x^n+y^n=z^n$ leads to $x=z$, which is absurd. So wlog. $x<y<z$. Note that $y^n=z^n-x^n$ a multiple of $z-x$, which must therefore be a power of $y$ (and $>y^0$). Thus $z\ge x+y$. But $(x+y)^n>x^n+y^n$.

Answer 2

It can be observed that if $k$ is prime, then either $k\equiv 1 \space(\text {mod 6})$ or $k\equiv 5 \space(\text {mod 6})$.

It is also observable that if $k\equiv 1 \space(\text {mod 6})$ or $k\equiv 5 \space(\text {mod 6})$, then $k^n\equiv 1 \space(\text {mod 6})$ or $k^n\equiv 5 \space(\text {mod 6})$.

Let us analyze $x$ and $y$ modulo $6$

Now as $x$ and $y$ are primes, both of them can either be congruent to $1$ modulo $6$ or congruent to $5$ modulo 6.

$\therefore x^n\equiv 1 \space(\text {mod 6})$ or $x^n\equiv 5 \space(\text {mod 6})$

Similarily $y^n\equiv 1 \space(\text {mod 6})$ or $y^n\equiv 5 \space(\text {mod 6})$

If both $x^n\equiv 1\space \text{(mod 6)}$ and $y^n\equiv 1\space \text{(mod 6)}$, then $(x^n+y^n=z^n)\equiv 2\space \text{(mod 6)}$ which means that $z$ is not prime.

Similarly if both $x^n\equiv 5\space \text{(mod 6)}$ and $y^n\equiv 5\space \text{(mod 6)}$, then $(x^n+y^n=z^n)\equiv 4\space \text{(mod 6)}$ which means that $z$ is not prime.

Finally if $x^n\equiv 1\space \text{(mod 6)}$ and $y^n\equiv 5\space \text{(mod 6)}$, then $(x^n+y^n=z^n)\equiv 0\space \text{(mod 6)}$, which means that $z$ again is not a prime.

We have proved that in any of the cases where $x$ and $y$ are primes, it is impossible for $z$ to be prime.

$\therefore$ There are no solutions for $x^n+y^n=z^n$ for $n\geq 2$ where $x, y$ and $z$ are primes

Answer 3

First, by Fermat's Last Theorem, this equation has no integral solutions, and thus no prime solutions, for all $n\geq 3$, so we only need to focus on the $n=2$ case.

Case 1: $x, y, z>2$

Because $x, y,$ and $z$ are and all primes greater than $2$, they are trivially odd. Hence, $x^2$, $y^2$, and $z^2$ are all odd. Adding $x^2$ and $y^2$ then yields an even number, but $x^2 +y^2 =z^2$, a contradiction.

Case 2: $x$ or $y=2$

WLOG, let $x=2$. Then, the equation yields $$2^2 +y^2=z^2.$$ Rearranging, $$y^2=z^2-2^2$$ $$\therefore y^2=(z-2)(z+2).$$ Because the prime factorization of $y^2$ is unique, the values $z-2$ and $z+2$ must be the products of prime factors (and possibly $1$) of $y^2$. However, because $y$ is the only prime factor of $y^2$, $y=z-2=z+2$, a contradiction.

Case 3: $z=2$

The only lattice points that the curve $x^2 +y^2 =2^2$ passes through are $(2,0)$, $(0,2)$, $(-2,0)$, and $(0,-2)$. The number $0$ is composite, so the equation does not have any solutions. $\blacksquare$

Answer 4

The three primes cannot all be odd, so one of them must be $2$. It cannot be $z$, so let's let it be $x$, in which case $y$ and $z$ are odd and we have

$$2^n=z^n-y^n=(z-y)(z^{n-1}+\cdots+y^{n-1})$$

which implies $z-y=2^k$ for some $1\le k$ (ruling out $k=0$ since $y$ and $z$ are odd). Writing $z=2^k+y$, we see that, since $n\ge2$, we have

$$2^n\ge2^k(2^k+y)^{n-1}=2^k(2^{k(n-1)}+\cdots+y^{n-1})\gt2^{kn}\ge2^n$$

which is a contradiction because of the strict inequality.

Answer 5

Reposting my answer from the deleted thread here by Martin's request.

Modulo two consideration shows that one of $a,b,c$ needs to be even. Given that there is a single even prime, we can conclude that $a$ or $b$ must be two (leaving the case $c=2$ as an exercise). W.l.o.g. $a=2$ and $b,c $ are both odd. But then $c\ge b+2$, and proving the inequality $$ 2^n+b^n<(b+2)^n $$ is another very simple exercise.

Observe that the arguments works even when $n=2$.

Answer 6

If $x$, $y$ and $z$ are primes with $x^n+y^n=z^n$ for an integer $n\ge 2$, then they can't all be odd, so at least one of them must be even. Since $2$ is the only even prime, at least one of them must be $2$. Since $2$ also is the smallest prime and $z$ must be greater than $x$ and $y$, $z$ can't be $2$, so w.l.o.g. $x=2$.

Now, since $z>y$ and $z\equiv{y}\pmod2$, we have $z \ge y+2$ and so finally we get a contradiction (because $n\ge2$): $$ 2^n=z^n-y^n\ge (y+2)^n-y^n=2^n+\sum_{k=1}^{n-1}\binom{n}{k}2^ky^{n-k}>2^n $$

Answer 7

Fermat's last theorem, now proven, shows that $A^x+B^x=C^x$ cannot be true for $x\ne2$ so it is not an issue here.

If we use a formula I developed that generates only the subset of Pythagorean triples where $(C-B)$ is an odd square, we can see some relationships that show how two numbers of a triple can be prime, but not all three.

$$A=(2n-1)^2+2(2n-1)k \qquad B=2(2n-1)k+2k^2 \qquad C=(2n-1)^2+2(2n-1)k+2k^2$$

The value of $C$ must be an odd number of the form $(C=4n+1)$ and and some of these numbers are prime but we can see that $A=(2n-1)^2+2(2n-1)k\implies A=(2n-1)(2n-1+2k)$ so $A$ is composite for all $(n>1)$–– only $(n=1)$ can yield prime numbers for both $A$ and $C$.

The value of $B$ can never be $2$ because the smallest is $4$ and it happens the $B$ is always a multiple of $4$ so $B$ cannot be prime.

$\therefore$ No Pythagorean triple, primitive or otherwise, can contain three prime numbers.