# Integer Parameterization of degree 2 equation: $(6x+5)y - x - 2 = (6w + 7)z - w + 4$

Finding a complete integer parameterization of $$(6x+5)y - x - 2 = (6w + 7)z - w + 4$$ has proved challenging. Can anyone lead me in the right direction? This is related to some nonprofessional research I am doing on the twin prime conjecture.

Goals: Find parameterizations for x, y, w, and z which keep each one an integer regardless of input and cover every situation in which the equality holds true.

Thank you!

• For what it's worth, virtually every MathSE posted question that I have seen, that followed this article on MathSE protocol has been upvoted rather than downvoted. I am not necessarily advocating this protocol. Instead, I am merely stating a fact: if you scrupulously follow the linked article, skipping/omitting nothing, you virtually guarantee a positive response. Commented Jul 17 at 19:20
• The degree of the polymials is $2$ , so in which sense is this a linear equation ? Commented Jul 17 at 19:21
• @Peter Ah, my bad, I was thinking linear just meant no variable is a square-- but you're right-- two variables multiplied is degree 2. I'll fix that. Commented Jul 17 at 19:48
• @user2661923 Is there anything I am missing? Commented Jul 17 at 19:50
• Please follow the instructions in my first comment. Commented Jul 17 at 19:51

Maybe this helps

$$(6x+5)y - x - 2 = (6w+7)z - w + 4$$

$$(6x+5)y - (x + 3) = (6w+7)z - (w - 3)$$

$$a = x+3,b = w-3$$

$$(6(a-3)+5)y - a = (6(b+3)+7)z - b$$ $$(6a - 13)y - a = (6b +25)z - b$$

$$6ay + b = 6bz + 25z + 13 y + a$$

$$(3a+y)^2 = 9a^2 + 6ay + y^2$$ $$(3b+z)^2 = 9b^2 + 6bz + z^2$$

$$(3a+y)^2 + 9b^2 + z^2 + b = (3b+z)^2 + 9a^2 + y^2 + 25 z + 13 y + a$$

This equation has many solutions. The sum of 3 squares is a very dense set.

We can even express variables in terms of each other. for instance $$a=b,y=z$$ reduces to

$$25 z + 13 y = 0$$

$$z = 25 k , z = -13 k$$

for integer $$k$$.

Or if $$a=y=0$$ we get

$$b = 6bz + 25z$$

$$0 = b(6z-1) + 25 z$$

and the only solutions of that are

$$b = -5,z=1$$ $$b=z=-4$$ $$b=z=0$$

So on the one hand we have many solutions, probably to many to fit in a single algebraic parameterisation. We can even write variables as functions of each other.

On the other hand some special cases only have a finite number of solutions.

So I suspect that there are $$2$$ or $$3$$ almost free variables and thus it is more like a relation than an equation.

Therefore I assume no closed form solution exists.

$$4$$ or $$5$$ variables is maybe surprisingly much.

Consider

$$(z + 2 - gh)(z - xy) = 0$$

this relates to prime twins if all variables are larger than 1. And I assume there is no parameterisation for $$z$$.

• Why the OP sees a connection to prime twins is unclear to me.
– mick
Commented yesterday