Prove that there exist $x_j, y_j, z_j ∈ A$ such that $a^{50}x_j +b^{20}y_j +c^{15}z_j = 1$.

Question

Let $A$ be a commutative ring with $1$, and let $a, b, c \in A$. Suppose there exist $x, y, z ∈ A$ such that $ax+by +cz = 1$. Then there exist $x_j, y_j, z_j ∈ A$ such that $a^{50}x_j +b^{20}y_j +c^{15}z_j = 1$.

I need some help with this question. Don't know where to start.

Answer 1

Hint: Expand $(ax+by+cz)^n=1$ for a sufficiently large $n$.

Answer 2

Here is a solution inspired by algebraic geometry: suppose, to the contrary, that we had $a^{50} x_j + b^{20} y_j + c^{15} z_j \ne 1$ for every $x_j, y_j, z_j \in A$. Then the ideal $\langle a^{50}, b^{20}, c^{15} \rangle$ is not the unit ideal, so it is contained in some maximal ideal $\mathfrak{m}$. However, then $a^{50} \in \mathfrak{m}$, and since a maximal ideal is prime, this implies $a \in \mathfrak{m}$. Similarly, $b \in \mathfrak{m}$ and $c \in \mathfrak{m}$ also. But now, $1 = ax + by + cz \in \mathfrak{m}$ also, giving a contradiction.