Solve a^2 + b^2 = c^3: Integer Solutions

[frax]

a^2 + b^2 = c^3
(a,b,c - integers, different from each other)
Is there a solution to this?

 

[Zurtex]

A trivial solution would be:

a = -1
b = 0
c = 1

 

[frax]

A trivial solution would be:

a = -1
b = 0
c = 1

Of course, for a,b,c you can have:
-5, 10, 5
-17, 68, 17
... and so on.
But let's add a condition |a|<>|b|<>|c|, or just
a, b, c - positive integers, different from each other.
Is there a solution, any idea?

 

[NateTG]

There's also
a=x^3,b=0,c=x^2

 

[arunbg]

Of course, for a,b,c you can have:
-5, 10, 5
-17, 68, 17
... and so on.
But let's add a condition |a|<>|b|<>|c|, or just
a, b, c - positive integers, different from each other.
Is there a solution, any idea?

I wrote a program in C++ for this and got the following results for
a,b,c upto 30

11, 2, 5
18, 26, 10

Of course u can put - signs for a and b.

 

[robert Ihnot]

For any prime congruent to 1 mod 4, it is the sum of two squares. For that matter so is the product of two such cases:

(a^2+b^2)(c^2+d^2) = (ac=+/-bd)^2+(ad-/+bc)^2.

This then can be carried to any degree. For example: 5=2^2+1^2 and fits the bill. We have:

5^3=5^2+10^2, and better: 2^2+11^2.

 

[frax]

 

[vaishakh]

Let there be a pythagorian triplet x^2 + y^2 = z^2. Now multiply both sides with z^4,
x^2*z^4 + y^2*z^4 = (xz^2)^2 + yz^2)^2 = z^6 = (z^2)^3.
Find individual pythagorian triplets and you will find corresponding solutions by this method.

 

[ramsey2879]

Let x^2 + y^2 = z. Now multiply both sides with z^2,
x^2*z^2 + y^2*z^2 = xz^2 + yz^2 = z^3