$a + b = c + c$ 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: 
(
 A: Suppose $c=u^2+v^2$ then you have $$(uc^2+v)^2+(vc^2-u)^2=c(c^4+1)$$ and you want $uc^2+v$ and $vc^2-u$ to be coprime.
Now just choose $u=1$ so that the expressions are $a=c^2+v$ and $b=vc^2-1$ with $c=v^2+1$
Note then that that a common factor of $a$ and $b$ is also a factor of $va-b=v^2+1=c$ and $ac^2-b=c^4+1$, but $c$ and $c^4+1$ are coprime.

So you get a family of solutions with 
$c=v^2+1$ and 
$a=c^2+v=v^4+2v^2+v+1$ and 
$b=vc^2-1=v^5+2v^3+v-1$
[Which, I notice is just Sam's parametric solution with $u=1$. It should be obvious what to do to generalise.]
A: Above equation shown below has parametric solution:
$(a^2+b^2)=c(c^4+1)$
$a=u^5+uv^4+2u^3v^2+v$
$b=v^5+u^4v+2u^2v^3-u$
$c=u^2+v^2$
For $(u,v)=(3,2)$  we get:
$(509^2+335^2)=13(13^4+1)$
A: 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:
$$Z=(a^2+b^2)^2$$
$$X=a^2+2(a^2+b^2)^4ab-b^2$$
$$Y=(a^2+b^2)^4a^2-2ab-(a^2+b^2)^4b^2$$
