Find the solutions of $(2a + b - y)(2x - b + y) - 2(a - x)(b - y) = (2c + b + y)(2z - b - y) - 2(c - z)(b + y)$

Question

Good day, I would like to know how would one find non-trivial positive integer solutions (if there are any) of the equation $(2a + b - y)(2x - b + y) - 2(a - x)(b - y) = (2c + b + y)(2z - b - y) - 2(c - z)(b + y)$, thank you.

Answer 1

There are non trivial solutions.

If you write $$f(a,b,c,x,y,z)=ax-ab+ay+bx+by-yx-cz+cb+cy-bz-yz$$

obtained by developing the equation, then you can try things like:

• $f(a,a,a,a,a,z)=0$ then $4a-3z=0$ so $a=3,z=4$ works
• $f(a,a,a,x,x,x)=0$ then $3a-2x=0$ so $a=2,x=3$ works also for instance.
• $f(a,b,c,a,b,c)=0$ then $a^2+b^2=c^2$ and you get all Pythagorean triplets (thus infinity of solutions).

However this is far from exhausting, I guess this is not easy to find everything...

Answer 2

Above equation is equivalent to:

$x(a+b)+y(a+b+c)-z(b+c)=xy+yz+b(a-c)$

For, $(a,b,c)=(1,2,3)$ we get:

$(3x+6y-5z)=(xy+yz-4)$

Above has solution:

$(x,y,z)=[(3k-5),(1),(k)]$

For, $k=4$ we get:

$(x,y,z,a,b,c)=(7,1,4,1,2,3)$

Answer 3

$(2a + b - y)(2x - b + y) - 2(a - x)(b - y)= (2c + b + y)(2z - b - y) - 2(c - z)(b + y)\tag{1}$
From equation $(1)$ ,we get x below $$x =\frac{ (-by-ay-cy+yz-cb+bz+cz+ab)}{(b+a-y)}$$ So, let $b+a-y=\pm1$ then we get the integer solution for $x$.

Substitute $y = a+b-1$ to equation $(1)$, then we get $$x = (a+2b-1+c)z-ab-b^2+b-a^2+a-ca-2cb+c.$$ Thus, we get a parametric solution.
$[x,y,z] = [(a+2b-1+c)z-ab-b^2+b-a^2+a-ca-2cb+c, a+b-1, z]$
$a,b,c,z$ are arbitrary.

Example:
$(a,b,c)=(1,2,3): [x,y,z]=[7z-16, 2, z]$
$(a,b,c)=(1,1,1): [x,y,z]=[3z-3, 1, z]$
$(z)=(a): [x,y,z]=[ab-b^2+b-2cb+c, a+b-1, a]$