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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!

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  • $\begingroup$ 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. $\endgroup$
    – user2661923
    Commented Jul 17 at 19:20
  • $\begingroup$ The degree of the polymials is $2$ , so in which sense is this a linear equation ? $\endgroup$
    – Peter
    Commented Jul 17 at 19:21
  • $\begingroup$ @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. $\endgroup$
    – ServingSpy
    Commented Jul 17 at 19:48
  • $\begingroup$ @user2661923 Is there anything I am missing? $\endgroup$
    – ServingSpy
    Commented Jul 17 at 19:50
  • $\begingroup$ Please follow the instructions in my first comment. $\endgroup$
    – user2661923
    Commented Jul 17 at 19:51

1 Answer 1

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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 $z$.

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

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