# Integer solutions to a^2+b^2=c^3

• Jun 11th 2009, 06:20 AM
linshi
Integer solutions to a^2+b^2=c^3
a^2+b^2=c^3, a,b,c are positive integers, and b>=a.

Just by inspection, a=b=c=2 is one set of solution.

Also, a=26, b=18, c=10; a=1358, b=594, c=130; are two other sets of solutions.

If you multiply a and b by n^3 and c by n^2 for any positive integers n, you can easily get related sets of solutions.

So my the question is:
- Do other solutions exists?
- Are there any solutions, such that a,b,c are pairwise coprime, exists?
• Jun 11th 2009, 07:48 AM
SimonM
One solution set is:

$\displaystyle (a,b,c) = (|x^3 - 3xy^2|,|3x^2y-y^3|, x^2+y^2)$ (and removing those where b<a)

Nor is it completely unmotivated. Consider the Gaussian integers:

$\displaystyle a^2+b^2 = c^3 \Rightarrow (a+ib)(a-ib) = c^3$

Suppose $\displaystyle a+ib = (x+iy)^3 = x^3 - 3xy^2 + i(3x^2y-y^3)$
• May 6th 2014, 07:08 AM
individ
Re: Integer solutions to a^2+b^2=c^3
And of course I wonder how she looks like a formula describing their solutions. For the special case when $\displaystyle X^2+Y^2=Z^3$

You can get a basic formula. Has the solutions:

$\displaystyle X=2k^6+8tk^5+2(7t^2+8qt-9q^2)k^4+16(t^3+2qt^2-tq^2-2q^3)k^3+$$\displaystyle 2(7t^4+12qt^3+6q^2t^2-28tq^3-9q^4)k^2+8(t^5+2qt^4-2q^3t^2-5tq^4)k+ \displaystyle +2(q^6-4tq^5-5q^4t^2-5q^2t^4+4qt^5+t^6) .................................................. .................................................. ............................................. \displaystyle Y=2k^6+4(3q+t)k^5+2(9q^2+16qt+t^2)k^4+32qt(2q+t)k^ 3+ \displaystyle 2(-9q^4+20tq^3+30q^2t^2+12qt^3-t^4)k^2+ \displaystyle +4(-3q^5-tq^4+10q^3t^2+6q^2t^3+5qt^4-t^5)k-$$\displaystyle 2(q^6+4tq^5-5q^4t^2-5q^2t^4-4qt^5+t^6)$

.................................................. .................................................. .............................................

$\displaystyle Z=2k^4+4(q+t)k^3+4(q+t)^2k^2+4(q^3+tq^2+qt^2+t^3)k +2(q^2+t^2)^2$

$\displaystyle q,t,k$ - What are some integers any sign. After substituting the numbers and get a result it will be necessary to divide by the greatest common divisor. This is to obtain the primitive solutions.

In the equation: $\displaystyle X^2+Y^2=qZ^3$ If the ratio is such that the root of an integer: $\displaystyle c=\sqrt{q-1}$ Then the solution is:

$\displaystyle X=-2(c+1)p^6+4(2c(q-2)-3q)p^5s+2(c(5q^2-2q-8)-q^2-22q+8)p^4s^2$$\displaystyle +8q(5q^2-14q+4)p^3s^3+2(c(5q^2-2q-8)+q^2+22q-8)q(q-2)p^2s^4- \displaystyle -4(2c(q-2)+3q)q^2(q-2)^2ps^5-2(c-1)q^3(q-2)^3s^6 .................................................. .................................................. .................................................. ..... \displaystyle Y=2(c-1)p^6+4(2qc+q-4)p^5s+2(c(-5q^2+18q-8)+15q^2-22q-8)p^4s^2$$\displaystyle +8q(q^2+6q-12)p^3s^3-2(c(5q^2-18q+8)+15q^2-22q-8)q(q-2)p^2s^4-$

$\displaystyle -4(2qc-q+4)q^2(q-2)^2ps^5+2(c+1)q^3(q-2)^3s^6$

.................................................. .................................................. .................................................. ......

$\displaystyle Z=2p^4+8p^3s+4(q^2-2q+4)p^2s^2-8q(q-2)ps^3+2q^2(q-2)^2s^4$

And more.

$\displaystyle X=-2(c-1)p^6-4(2c(q-2)+3q)p^5s+2(c(5q^2-2q-8)+q^2+22q-8)p^4s^2$$\displaystyle +8q(5q^2-14q+4)p^3s^3+2(c(5q^2-2q-8)-q^2-22q+8)q(q-2)p^2s^4+ \displaystyle +4(2c(q-2)-3q)q^2(q-2)^2ps^5-2(c+1)q^3(q-2)^3s^6 .................................................. .................................................. .................................................. ....... \displaystyle Y=2(c+1)p^6-4(2qc-q+4)p^5s+2(c(-5q^2+18q-8)-15q^2+22q+8)p^4s^2$$\displaystyle +8q(q^2+6q-12)p^3s^3+2(c(-5q^2+18q-8)+15q^2-22q-8)q(q-2)p^2s^4+$

$\displaystyle +4(2qc+q-4)q^2(q-2)^2ps^5+2(c-1)q^3(q-2)^3s^6$

.................................................. .................................................. .................................................. .......

$\displaystyle Z=2p^4-8p^3s+4(q^2-2q+4)p^2s^2+8q(q-2)ps^3+2q^2(q-2)^2s^4$
• May 6th 2014, 08:22 AM
Idea
Re: Integer solutions to a^2+b^2=c^3
Quote:

Originally Posted by linshi
a^2+b^2=c^3, a,b,c are positive integers, and b>=a.

Just by inspection, a=b=c=2 is one set of solution.

Also, a=26, b=18, c=10; a=1358, b=594, c=130; are two other sets of solutions.

If you multiply a and b by n^3 and c by n^2 for any positive integers n, you can easily get related sets of solutions.

So my the question is:
- Do other solutions exists?
- Are there any solutions, such that a,b,c are pairwise coprime, exists?

A pairwise coprime solution

$\displaystyle 2^2+11^2=5^3$