The diophantine equation $ x^p+y^p=kz^p$ has infinitely many solutions?

Question

Is that true that the diophantine equation $ x^p+y^p=kz^p$ has infinitely many nontrivial solutions when $p,x,y,z,k>2, gcd(x,y,z)=1$ ? If so, how do you prove it?

Answer 1

Just asked Pari/GP for some solutions and got this:

3^3+5^3=2^3*19
3^3+13^3=2^4*139
3^3+29^3=2^5*763
3^3+37^3=2^3*6335
3^3+53^3=2^3*18613
3^3+61^3=2^6*3547
3^3+77^3=2^4*28535
3^3+85^3=2^3*76769

It seems, that is an infinite family of solutions; we have $$ 3^3 + (5+8k)^3 = 2^3 \cdot x \qquad \qquad k \in \{0,1,3,4,6,7,9,10,... \}$$ and might be better written as $$ 3^3 + (8+24 k-3)^3 = 2^3 \cdot x \qquad \qquad k \ge 0 \tag 1$$ $$ 3^3 + (16+24 k-3)^3 = 2^3 \cdot x \qquad \qquad k \ge 0 \tag2$$ The first one gives: $$ 8^3(1+3 k)^3 -9\cdot 8^2(1+3k)^2+27\cdot 8 (1+3k) = 2^3 \cdot x_k \tag 3 $$ which is obviously true since rhs and lhs have the factor $2^3$ in common. Obviously this is the same for the second form.

So we have already one infinite family of solutions.

---

[update] In the light of my comment above here are some less trivial solutions:

8^3+199^3=7^3*22977
8^3+495^3=7^3*353609
8^3+885^3=7^3*2020859
8^3+1181^3=7^4*686053
8^3+1571^3=7^3*11304061
8^3+1867^3=7^3*18973125
...
8^3+1705^3=13^3*2256021
....
8^3+1983^3=19^3*1136861

and it seems this would continued if I would increase the second base above 2000 .

[end update]

---

---

Additional remark: looking from the other side at the problem, I've seen in R. Guy's "unsolved problems..." (1994) that statement:

"(...) which shows the unsolvability of $x^p+y^p+\gamma z^p=0$ if $p \gt 11$ and $\gamma$ is the power of one of the primes $3,5,7,11,13,17,19,23,29,53,59,\ldots$ . (...)"

.... which made me the completely wrong impression about the question. Upps...

Answer 2

Hint: There are lots of solutions with $z=1$.