Prove that the following is not true or true. Prove that the following is not true or true. If $x,y,z \geq 1$ and they are all integers then $z^2=9xy+6y+6x+3$. If true, find $x,y,z$. If false, prove it.
I first tried some values but the fact that I am being asked to prove it if it is false, then I guess the above is false.
I tried some values like
$x=1,y=1$ then $z^2=24$ but that violates $z$ being an integer.
So if the above is false, how would i proceed in proving it.
EDIT: Sorry for all the confusion but I'm not trying to approach this problem by counterexamples. I want to know why does the equation $z^2=9xy+6y+6x+3$ not work? Why wont the left hand side ever provide a solution on the right hand side
 A: So, there are two possible questions here: 
$1)$ As stated, your question states, prove or disprove that $\forall x,y,z\in \mathbb{Z}_{\ge 1}$ 
$$ z^2=9xy+6y+6x+3.$$
Here is the counterexample: consider $x=y=z=1$. This implies
$$ 1=24.$$
This is blatantly false, and so this can not hold for all $x,y,z,\in \mathbb{Z}_{\ge 1}$.
$2)$ The alternative question states, prove or disprove that $\exists x,y,z\in\mathbb{Z}_{\ge1}$ such that 
$$z^2=9xy+6y+6x+3. $$
Through software, or some sort of exhaustive calculation, we can see that $x=5,y=8\implies z=21$, a solution to the equation. So there exists such a $3-$tuple in $\mathbb{Z}_{\ge1}$ which solves the equation.
A: If you reformulate your statement as (which I guess is what the question is asking)
There doesn't exist integers $x,y,z≥1$ that satisfies $z^2=9xy+6y+6x+3$
Then the statement is still false. An counterexample would be $(x,y,z)=(1,9,12)$.
(It's easy to see $3|z$, and you can rewrite the equation as $z^2+1=(3x+2)(3y+2)$, so we just need to find a number of the form $(3m)^2+1$ that can be factored into two $3n+2$ type numbers. Trying out the first few possibilities gives $12^2+1=145=5\times29$.)
A: If it is not true then there exists a counterexample. You just found one. You're done.
A: The way the statement is formulated, you can pick $x,y$, and $z$ all at the same time, then check whether equality holds. For instance, choosing $x=y=z=1$ gives:
$$1=24$$
which is absurd. This is enough to prove that the statement is false.
Also note that it doesn't really make sense to write "if true, find $x,y,z$", because the main question, as written, is whether the equation holds for all possible $x,y,z$.
A: Here is a solution: $x=5,$ $y=8,$ $z=21.$
First note that $z$ must be a multiple of 3 and put $z=3k.$ Then note that $x$ and $y$ must be congruent to 2 modulo 3 and write $x=3l-1,$ $y=3m-1.$ Now the equation simplifies to
$$k^2=9lm-l-m.$$
Just try a few values of $l$ and $m$; the first ones that work are 2 and 3.
A: You haven't quite understood the question.
z^2 is defined as the result got by adding together 9xy + 6y + 6x +3 
you're being asked whether 9xy + 6y + 6x + 3 can be a perfect square,
when x and y are both positive integers.
Many integers are not squares; they want you to show 9xy + 6y  + 6x + 3 can be a square, presumably for just one set of values, though this seems unclear.
One approach might be to try to find x and y so that 3xy + 2y + 2x +1 is 3 x a square, such as 3, 12, 27, 48... maybe by factorising.
(x + 1)(y+1) = xy + y + x +1 umm..
or let's try x=y, so we get 9X^2 + 12x + 3 = 3(3x+1)(x+1).
Then say x+1 = 3M to allow for the 3
= 3(3M)(3x + 1)  to get a square, M (3x+1) must be square (where M=(x+1)/3)
so M( 3 (3M -1) + 1) has to be a square.
so M(9M -2) has to be square. So ... ummm...
Anyway, give it a go....
