# Find $a$ and $b$ with whom this expression $x\bullet y=(a+x)(b+y)$ is associative

I need to find a and b with whom this expression is associative: $$x\bullet y=(a+x)(b+y)$$ Also known that $$x,y\in Z$$ So firstly I write: $$(x\bullet y)\bullet z=x\bullet (y\bullet z)$$ Then I express and express them and after small algebra I get: $$ab^2+b^2x+az+abz+bxz=a^2b+a^2z+bx+abx+axz$$ And I don't know what to do now, I don't know if I should do it, but I tried to assign to $x=0$ and $z=1$ but nothing better happened.

Could you please share some ideas?

EDIT: Sorry, but I forgot to mention that I need to find such a and b with whom it should be associative

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Unless there's something special about $a$ and $b$, your operator isn't associative... – J. M. Oct 16 '11 at 16:21
@J.M. sorry, forgot to add that I need to find such a and b with whom it should be associative – Templar Oct 16 '11 at 16:34
Obviously, $a=b=0$ works. – J. M. Oct 16 '11 at 16:36
@Templar: What are $a,b,x,y$ supposed to be? Real numbers? Integers? – Zev Chonoles Oct 16 '11 at 16:37
@ZevChonoles x,y is integers, nothing said about a, b – Templar Oct 16 '11 at 16:40

It's associative if $a=b=0$. I looked at $a=b=1$ and $a=0,\ b=1$ and ruled those out. If you want coefficients of $z$ to match, you need $a^2=a$, and that happens if and only if $a=0$ or $a=1$, and similar comments apply to $b$.

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But how to prove it? Or it's enough just try $a=b=0$; $a=b=1$; $a=0$, $b=1$; $a=1$, $b=0$? – Templar Oct 16 '11 at 16:45
Look at the identity $ab^2 + b^2x + az + abz + bxz = a^2b + a^2z + bx + abx + axz$. The coefficient of $x$ on the left side is $b^2$, and on the right side is $b+ab$. So we get $b^2 = b+ab$, and several other identities derived similarly. There's also the question of why two polynomials can be equal only if their corresponding coefficients are equal---one could go into how to prove that. – Michael Hardy Oct 16 '11 at 17:42

Consider $a$ and $b$ to be fixed. To be associative, your association equation must work for all $x, y, z \in \mathbb{Z}$. Start with $x = y = z = 0$. Then what you have would simplify to $a^2b = ab^2$. The solution set is all pairs $(0, b)$, all pairs $(a, 0)$, and all pairs for which $a=b$ (this we get by considering each of $a$ and $b$ nonzero and zero separately).

Now, in the case $a = 0$, the association equation reduces to $$b^2x + bxz = bx$$ Clearly, if $b=0$ then the association works, so assume $b \neq 0$. Dividing by $b$:

$$bx + xz = x$$ $$(b + z)x = x$$

But this cannot work for arbitrary $x, z$ (consider $x = 1$ and $z = b$). In a similar way, in the case $(a, 0)$, we can rule out nonzero $a$.

Finally, consider the case $a=b$. Plug in $a$ instead of $b$ in the association equation, and after all the dust settles, you are left with $az = ax$. If $a \neq 0$, then this would imply $z=x$, not true for arbitrary $x, y, z \in \mathbb{Z}$. Thus, the only possibility that remains is $(a, b) = (0, 0)$.

Hope this helps!

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You can rewrite your condition as $$(ab^2 - a^2b) + (b^2 - b - ab)x + (a + ab - a^2)z + (b-a)xz = 0$$ This has to be true for all integers $x$ and $z$. This means each of the four coefficients above has to be zero; there is only one polynomial which is zero at every $(x,z)$ for which $x$ and $z$ are both integers. (To see this directly in your case: set $x = 0$; this gives that the first and third cofficients are zero. Then set $z = 0$; this gives that the second coefficient is zero. Finally set $x = z = 1$ to get the final one to be zero.)

Since the $xz$ coefficient is zero, $a$ has to equal $b$. This also makes the constant term zero. Then the second and third coefficients being zero reduce to $b = 0$ and $a=0$ respectively. Thus $a=b=0$ is the only way to make the operation associative.

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Assuming that you still want the element $1 \in \mathbb{Z}$ to be a multiplicative identity element with your new multiplication, here is an alternative way of quickly seeing that $a=b=0$.

For an arbitrary $y\in \mathbb{Z}$ we shall have:

$y=1 \bullet y=(a+1)(b+y)=ab+ay+b+y=(a+1)b+(a+1)y$

This is satisfied if and only if $a+1=1$ and $(a+1)b=0$. The unique solution is $a=0$ and $b=0$.

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