$a + b = c + c$ So I have the following problem: $a + b = c + c.
I want to prove that the equation has infinitely many relatively prime integer solutions.
What I did first was factor the right side to get: 
(

In the equation:
$$X^2+Y^2=Z^5+Z$$
I think this formula should be written in a more general form:
$$Z=a^2+b^2$$
$$X=a(a^2+b^2)^2+b$$
$$Y=b(a^2+b^2)^2-a$$
And yet another formula:
$$Z=\frac{a^2+b^2}{2}$$
$$X=\frac{(a-b)(a^2+b^2)^2-4(a+b)}{8}$$
$$Y=\frac{(a+b)(a^2+b^2)^2+4(a-b)}{8}$$
$a,b$ - arbitrary integers.
Solutions can be written as follows:
$$Z=\frac{(a^2+b^2)^2}{2}$$
$$X=\frac{((a^2+b^2)^4+4)a^2+2((a^2+b^2)^4-4)ab-((a^2+b^2)^4+4)b^2}{8}$$
$$Y=\frac{((a^2+b^2)^4-4)a^2-2((a^2+b^2)^4+4)ab-((a^2+b^2)^4-4)b^2}{8}$$
where $a,b$ - any integers asked us.
Well, a simple solution:
$$Z=(a^2+b^2)^2$$
$$X=a^2+2(a^2+b^2)^4ab-b^2$$
$$Y=(a^2+b^2)^4a^2-2ab-(a^2+b^2)^4b^2$$
