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The question is from Zeitz’s ”The Art and Craft of Problem Solving:”

Find all positive integer solutions $x,y,z,p$, with $p$ a prime, of the equation $x^p + y^p = p^z$.

One thing I noticed is that

- A diophantine equation
- Show that if $p$ and $q$ are primes $\equiv 3$(mod $4$) then at least one of the equations $px^{2}-qy^{2} = \pm 1$ is soluble in integers $x, y$.
- Solve the Diophantine equation $ 3x^2 - 2y^2 =1 $
- Solving: $3^m-2=n^2$
- Generate solutions of Quadratic Diophantine Equation
- No integer solutions for $x^5 - 3y^5 = 2008$

$$ \frac{x^p + y^p}{x + y} = \sum_{i=0}^{p-1}x^{(p-1)-i}(-y)^i \implies (x+y) |p^z \implies x + y = p^n, \text{ }n < z $$

It is also not hard to determine all solutions for $p =2$. After this, however, I am at a loss. I don’t know what restrictions I can impose to try to narrow down the solution set; for example, the class of solutions

$$ p=3, x = 3^n, y = 2\cdot3^n, z = 2 + 3n $$

for $n \ge 0$ show that $x+y = p^n$ cannot be sharpened. (Let me note that these are the only other solutions I have found besides the ones for $p = 2$.)

Could anybody give me a push in the right direction on this problem?

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- Prove that there does not exist integer solutions for the diophantine equation $x^5 - y^2 = 4$.

Sorry, I can’t think of a good hint to give you a push. Here is my solution. There likely is a better way of approaching it, since this is from Art and Craft.

Deal with the case $p=2$ separately. Henceforth, $p$ is an odd prime.

If $\gcd (x, y) = k > 1$, then $k^p \mid p^z$, which implies that $k \mid p^z$. Thus, we can divide out by $k$. Henceforth, assume that $\gcd(x,y) = 1$.

Since $x + y \mid x^p + y^p$, hence $ x+y = p^n$. We have

$$x^p + (p^n – x)^p = p^z.$$

With the condition that $\gcd(x,y) = 1$, we get that $ p \not \mid x$. As such, $p^{n+1}$ divides $x^p + (p^n – x)^p$ but $p^{n+2}$ doesn’t. Hence,

$x^p + (p^n – x)^p = p^{n+1} $.

This becomes extremely restrictive. For $p \geq 5$, we have

$LHS \geq \frac{p^{5n}}{16} > p^{n+1}=RHS$,

Hence, the only possibility is $p=3$.

So this is very easy to solve if you just use the LTE Lemma.

$z=v_p(p^z)=v_p(x^p+y^p)=v_p(x+y)+v_p(p)=v_p(x+y)+1 \Rightarrow v_p(x+y)=z-1$ so we can write that $x+y=p^{z-1}\alpha$, where alpha is a positive integer which is not divideable with p.

It’s easy to show that $\alpha=1$ bcs we have $(x+y)(x^{p-1}-…+y^{p-1})=p^{z-1}\alpha (x^{p-1}-…+y^{p-1})=p^z$, and bcs $p\nmid\alpha$ we have that $\alpha=1$.

This everything works only for odd p.

Now for $p\ge5$ we have:

$x^p+y^p \ge 2(\frac{x+y}2)^p>p(x+y)$

$(\frac{x+y}2)^{p-1}>2^{p-1}$, bcs of $x+y=p^{z-1}\ge5^{z-1}>4^{z-1}\ge4$

And ofc $2^{p-1}$ growes faster than $p$ for $p\ge5$ so $2^{p-1}>p$.

Now you just have to see for $p=2$ and $p=3$ and that is it.

There exist more elementary solutions, but Zsigmondy’s theorem solves it.

$a^p+b^p=p^c$ has no solutions ($a,b,c\ge 1$, $p$ prime), except for $$(\{a,b\},c,p)=(\{2,1\},2,3),(\{2^k,2^k\},2k+1,2),\,k\in\Bbb Z_{\ge 0}$$

- $a=b$. Then $(a,b,c,p)=(2^k,2^k,2k+1,2)$, $k\in\Bbb Z_{\ge 0}$.
- $p=2$. Let $(a,b)=(2^ka_1,2^kb_1)$, $k\ge 0$ and $a_1,b_1\ge 1$ odd. $a_1^2+b_1^2=2^{c-2k}$.

$4\mid a_1^2+b_1^2\,\Rightarrow\, 2\mid a_1,b_1$, so $c-2k\in\{0,1\}$, $(a,b,c,p)=(2^k,2^k,2k+1,2)$. - wlog $a>b$, $p>2$. $(a,b,c,p)=(2,1,2,3)$, otherwise by Zsigmondy’s theorem $$a^p+b^p=(a+b)(a^{p-1}-a^{p-2}b+\cdots+b^{p-1})$$ has a prime divisor that does not divide $a+b\ge 2$, so has at least two prime divisors.

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