I have a result of an integral where I need to be able to find integer solutions once a C is specified:
$$F(a,b,z,y) = abzy + (a+y) \cdot (b+z) + C$$
My idea is as follows: find a $C'$ that allows the integral above to be written as a single product (and/or quotient). Then once the real $C$ is specified, factor $C-C'$, and match factors with the product above.
The problem is, how to obtain a factorization of the above as a single factor. With an appropriate $u$-substitution, we can introduce a constant $D$ in there, which might help recognize a factorization; also, modifying the $C$ a bit seems to help:
$$F(a,b,z,y)+C=abzy+Dby+Daz+ab+1+D^2$$
I will give upvotes even for progress toward a solution, e.g. something I haven't tried. Once I go down the wrong path once, I always wanna come back to that path. So since what I tried didn't work, I recommend not reading things I've tried until you thought a bit about it yourself. Therefore, here is the outline of the remainder of what I've tried:
- Part 1
- Hint for Part 2
- Part 2
- Hint for Part 1
In other words, if you only want a hint for Part 1, you would read at the bottom. So here goes.
Part 1:
Multiply everything by the factor $E=D$. Using $u$-substitution again, we can write $u=ab+1,v=zy+1$, to obtain
$$D\cdot F(a,b,z,y)+C=uv+Dby+Daz+D^2$$
Again, at this point I hope multiplying everything by another factor $E$ will help. However, I received nothing better than a sum of products. Namely, when $E=u$ I obtained $\color{gray}{u\cdot D\cdot F(a,b,z,y)+Cu{\rm\ equal\ to:}}$
$$(uz+Db)(uy+Da)-D^2ab+u^2+D^2u=(uz+Db)(uy+Da)+u^2+D^2$$
The last simplification was obtained by expanding the definition of $u$. With an appropriate change of $C$, we can simplify it down to $\color{gray}{u\cdot D\cdot F(a,b,z,y)+C'u{\rm\ equals:}}$
$$(uz+Db)(uy+Da)+(u-D)^2$$
So this does not work.
Now, the expression is perfectly symmetrical in $\{z,y\}$ and $\{a,b\}$. In the sense that if I multiply by $E=v$ instead, I will eventually get the same result: $\color{gray}{v\cdot D\cdot F(a,b,z,y)+Cv{\rm\ equals:}}$
$$(vb+Dz)(va+Dy)+(v-D)^2$$
One other idea I had -- what if I multiply by both $u$ and $v$? In that case, $\color{gray}{uv\cdot D\cdot F(a,b,z,y)+Cuv{\rm\ will\ equal:}}$
$$(uv+Dby)(uv+Daz)-D^2abzy+D^2uv=(uv+Dby)(uv+Daz)+D^2(u+v+1)$$
Again, there is a second factor present.
Hint for Part 2: At this point I realized that we can do further $u$-substitution, to try to make the factor of $(u-D)$ disappear.
Part 2:
I tried $u$-substitution $\alpha=a\sigma$ and $\beta=b\sigma$, with $\sigma$ to be determined.
The result was that $\color{gray}{D\cdot\sigma^2\cdot F(a,b,z,y)+C\sigma^2\rm{\ was\ equal\ to}}$: $$(\alpha\beta+\sigma^2)(zy+1)+D\sigma\beta y+D\sigma\alpha z+(D\sigma)^2$$
This time the $u$-substitution was, that $u=\alpha\beta+\sigma^2$, with $v=zy+1$ as before.
The result was that $\color{gray}{D\cdot\sigma^2\cdot u\cdot F(a,b,z,y)+C\sigma^2{\rm\ was\ equal\ to}}$:
$$(uz+D\sigma\beta)(uy+D\sigma\alpha)+(D\sigma^2)(u-\alpha\beta)+u^2=(uz+D\sigma\beta)(uy+D\sigma\alpha)+(D\sigma^2)(\sigma^2)+u^2 $$
Again by adjusting the $C$, we have $\color{gray}{D\cdot\sigma^2\cdot u\cdot F(a,b,z,y)+C'\sigma^2{\rm\ equal\ to}}$ $$(uz+D\sigma\beta)(uy+D\sigma\alpha)+(u-D\sigma^2)^2=(uz+D\sigma\beta)(uy+D\sigma\alpha)+(\alpha\beta+(1-D)\sigma^2)^2 $$
In other words, if I want the extra factor to disapear, then I need to set $\sigma=\sqrt{\frac{\alpha\beta}{D-1}}$. But, if I make that substution, then $\alpha$ and $\beta$ end up being related by $\alpha=\frac{a^2}{D-1}\beta$ and $\beta=\frac{b^2}{D-1}\alpha$. Therefore, this ends up imposing restrictions on (the variables) $a$ and $b$, instead of actually factoring this.
Is it possible that no factorization exists??
Hint for Part 1: use a "integration factor" $E$ type of idea, to write $E \cdot F(a,b,z,y)$ as a single product.