# Field theorem: impossible to satisfy three equations simultaneously in an integer field?

Suppose $a_0,a_1,b,c,d$ with $a_0 \neq a_1$ are integers, or elements of any ${\mathbb Z}_p$ for prime $p$.

A while ago I was attempting to find a set of such numbers that satisfied the equations

$$a_0 a_0 = ba_0 + ca_0 + d$$ $$a_0 a_1 = ba_0 + ca_1 + d$$ $$a_1 a_1 = ba_1 + ca_1 + d$$

and after many failed attempts ultimately conjectured that it was impossible.

I set about trying to prove this, but was immediately swamped in equations and ring axioms, none of which availed any insights.

I also noted that the equations can be satisfied in integer rings. For example, $a_0 = 2, a_1 = 5, b,c = 8, d = 8$ in ${\mathbb Z}_9$.

Can anyone provide insights as to how to prove the conjecture? Or is it in fact wrong, and can somebody procure a set of numbers that satisfies the equations in the specified fields?

Many thanks for any assistance.

Edit: I should note that the original problem I looked at had the additional condition that $$a_1 a_0 = ba_1 + ca_0 + d$$ which trivially requires that $b = c$.

This case is also handled by the supplied answer, however.

• $a_0 = 1 = c,\ a_1 = 0 = b = d\ \$ – Bill Dubuque Aug 14 '14 at 4:18

Assume we are in an integral domain.

If we subtract the first and third equations, we get $a_0^2 - a_1^2 = b a_0 + c a_0 - b a_1 - c a_1$, which is the same as $(a_0 - a_1)(a_0 + a_1) = (b + c)(a_0 - a_1)$. Since $a_0 \ne a_1$, $a_0 - a_1 \ne 0$, and so we can cancel to get $a_0 + a_1 = b + c$.

Subtracting the second and third, we get $a_0 a_0 - a_0 a_1 = c a_0 - c a_1$, which, after cancellation, becomes $a_0 = c$. This tells us that $a_1 = b$.

Now our equations have fewer variables, and we can see the redundancy: $$c^2 = bc + c^2 + d$$ $$bc = bc + cb + d$$ $$b^2 = b^2 + cb + d$$

All three tell us that $d = -bc$. The solution set can be parameterized as $(a_0, a_1, b, c, d) = (t, u, u, t, -tu)$. As an example, $(1, 2, 2, 1, -2)$ works:

$$1^2 = 2 \cdot 1 + 1 \cdot 1 + (-2)$$ $$1 \cdot 2 = 2 \cdot 1 + 1 \cdot 2 + (-2)$$ $$2^2 = 2 \cdot 2 + 1 \cdot 2 + (-2)$$

• That would make sense then. I originally had the requirement that $b = c$, meaning the solution would be $\left( t, t, t, t, -t^2\right)$, which violates the $a_0 \neq a_1$ rule. – COTO Aug 14 '14 at 4:41