Find the least positive integral value of z for which there exists different positive integers $x$,$y$ and $z$ that satisfy $x^2 +y^2=z^3$.

Question

My approach is as follows $(x+y)^2-z^3-2xy=0$ and by applying quadratic formula i get $z^3 + 2xy >0$ but now i get stuck.

Answer 1

Do you know the sum of two squares theorem? You can just start trying values for $z$. $1$ doesn't work because one of $x,y$ must be zero. $2$ doesn't work because the only solution is $x=y=2$ but you demand $x \neq y$. What about $z=3$? Before long you will find one that works, then you need to show that nothing lower does.

Answer 2

It's easy to brute-force (using a bit of Python code) that the smallest such $z$ is $z=5$, which satisfies the equation $5^3 = 2^2 + 11^2$.

import math
import sys

for i in range(10000):
    j = 1
    while j**2 < i**3:
        if abs((i**3 - j**2)**(1/2) - math.floor((i**3 - j**2)**(1/2))) < 0.001:
            print(i + "^3 = "+ j + "^2 + " + math.floor((i**3 - j**2)**(1/2)) + "^2")
            sys.exit()
    j += 1

Answer 3

squares mod 9:

0,1,4,0,7,7,0,4,1,...

cubes mod 9:

0,1,8,0,1,8,0,1,8....

possible sums:

0+0=0 0+1=1 1+7=8 4+4=8

which allows you to search 23 out of 64 possibilities. simply from there being 3 variables of positive value, we get z>2. Furthermore if z is odd, then one of the bases of the squares is, otherwise they are both even. I'll leave it to you to apply all this.