Does the equation, [tex: x^2 + y^3 = z^4] have solutions in prime numbers? Find at least one if yes, give a nonexistence proof otherwise.

from Vlad Mitlin via https://www.math.utah.edu/~cherk/puzzles.html

is i a prime? if so
x = i
y = 1
z = 0

Taking both sides of the equation modulo 4 and taking into account any odd prime modulo 4 is either 1 or 3, one has the following possibilities for the left hand side:
1^2+1^3 = 2 mod 4
1^2+3^3 = 0 mod 4
3^2+1^3 = 2 mod 4
3^2+3^3 = 0 mod 4

And for the right hand side:
1^4 = 1
3^4 = 1

So it's not possible for both primes odd.
There's still the case where either x=2 or y=2 that needs to be checked.

P9470
I did it in a very convoluted way, since the power of any odd number is odd, and the sum of 2 odd numbers is even, meaning that doing it mod 2 would have sufficed.

P9472
yea the proof is ridiculously simple when you look at like that
for numbers different than 2 the left side would necessarily be even and the right necessarily odd
but if any of then is 2, the problem is split in 3:
1. [tex: 4 + y^3 = z^4]
2. [tex: x^2 + 8 = z^4]
3. [tex: x^2 + y^3 = 16]
and its obvious that no other number can be 2 as well
so it is somewhat evident that 3. has no solution, but you can easily apply brute force to be sure. only 2 has a cube smaller than 16, and only 2 and 3 have squares smaller than 16, so the only combination possible is [tex: 3^2 + 2^3 = 9 + 8 = 17] which comes close but still no
1. and 2. are left as an exercise for the reader :^)

hey you can turn those into functions
1. [tex: 4 + y^3 = z^4 => f(y) = \sqrt[4]{y^3 + 4}]
or [tex: 4 + y^3 = z^4 => f(z) = \sqrt[3]{z^4 - 4}]
2. [tex: x^2 + 8 = z^4 => g(x) = \sqrt[4]{x^2 + 8}]
or [tex: x^2 + 8 = z^4 => g(z) = \sqrt[2]{z^4 - 8}]
so if you know of any tricks to know whether a function has solutions in the set of primes, thats gonna solve it

just finished brute testing those bitches for the first million primes
no solutions found
so im gonna ahead and do the applied math's chad move of saying that for all intents and purposes the solution is non existent
pure math cucks will seethe

P9516
now solve pic related ("exact" means rational)

i get to
[tex: c^3 + d^3 = 9]
where c and d are the circumferences of 2 phials, and 1 and 2 are the only integer solutions
as for rational solutions, idk

P9474
For the case 1, one could write the equation as
y^3 = z^4-4 = (z^2+2)(z^2-2)

Since y is prime one of the following must be true:
z^2+2 = 1 and z^2-2 = y^3
z^2+2 = y and z^2-2 = y^2
z^2+2 = y^2 and z^2-2 = y
z^2+2 = y^3 and z^2-2 = 1

The first and fourth cases lead to z = sqrt(-1) or z = sqrt(3), which aren't valid.
The second and third cases lead to a biquadratic equation that gives non-integer solutions as well.

P9660
And for case 2, one could write the equation as
8 = z^4-x^2 = (z^2+x)(z^2-x)

As before, one of the following must be true:
1 = z^2+x and 8 = z^2-x
2 = z^2+x and 4 = z^2-x
4 = z^2+x and 2 = z^2-x
8 = z^2+x and 1 = z^2-x

The first and fourth one lead to z = 3*sqrt(2)/2, which isn't valid.
The second and third one lead to z = sqrt(3), which isn't valid.

P9667
Looks solid.

P9474
Another way to go forward is to look at the equations modulo other numbers. A non-multiple of 3 is either 1 or 2 modulo 3, so its square is 1 modulo 3. So if neither x and z are multiples of 3, then [tex: y^3 = z^4 - x^2] is 1 - 1 = 0 modulo 3. Thus one of the numbers x, y, z must be 3. It's then easy to churn through all 6 possible positions of 2 and 3 and see that none of them produce an integer solution for the remaining variable.

>>P9660
WHY DOES SHE HAVE SIDEBURNS? WOULD YOU ***** A NECKBEARD?