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if $A=\{z\in \mathbb{Z}^+:z= 6xy +x-y; x\in \mathbb{Z}^+ ,y \in \mathbb{Z}^+\}$ what is $A^c$ in $\mathbb{Z}^+$? $A^c$ should have an equation form, without $\neq$ sign.

I make a transformation $6x(x+k)+k=6x^2+(6x+1)k$ and $6x(x-k)-k=6x^2+(6x-1)k$. But couldn't make to much progress. If you can help, I will be glad thanks.

Aylean Clara Grandieur
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$z=6xy+x-y\iff 6z-1=(6x-1)(6y+1). $

Now if $x,y\in \Bbb Z^+$ then $6x-1\ge 5$ and $6y+1\ge 7$ so $6z-1$ cannot be prime.

But if $6z-1$ is composite then it has a factor $a\equiv -1 \mod 6$ with $1<a<6z-1.$ So let $a=6x-1$ with $x\in \Bbb Z^+.$ Then $1<(6z-1)/a=b\in \Bbb Z^+$ with $b\equiv 1 \mod 6.$ So let $b=6y+1 $ with $y\in Z^+.$ Then $6z-1=(6x-1)(6y+1)$ so $z\in A.$

So $A=\{z\in \Bbb Z^+: 6z-1 \text { is not prime}\}.$

DanielWainfleet
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  • Nice solution! I only have a question about "if $6z-1$ is composite then it has a factor $a\equiv -1$" - I can check by inspection that if $xy\equiv 5,\mathrm{mod},6$ then either $x$ or $y$ is congruent $5$ modulo 6, but is there some more general statement that this follows from? – J_P Jul 13 '19 at 18:03
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    Thanks I think I get it. So $A^c={z\in \Bbb Z^+: 6z-1 \text { is prime}}.$ – Aylean Clara Grandieur Jul 13 '19 at 18:08