$a + b = c + c$

Question

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: (

Answer 1

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.

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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.]

Answer 2

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:

$(509^2+335^2)=13(13^4+1)$

Answer 3

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:

$$Z=(a^2+b^2)^2$$

$$X=a^2+2(a^2+b^2)^4ab-b^2$$

$$Y=(a^2+b^2)^4a^2-2ab-(a^2+b^2)^4b^2$$