How to prove that there exist α, β, γ ∈ R such that $αa^{15} + βb^{16} + γc^{17} = 1.$

Question

Let R be a commutative ring with unity. Let $ a, b, c ∈ R $ be such that there exist $x, y, z ∈ R$ with $xa + yb + zc = 1$

Show that there exist α, β, γ ∈ R such that

$$αa^{15} + βb^{16} + γc^{17} = 1.$$

MY trial : I take $αa=βb= γc= \frac{1}{3}$ where $a= b= c=1$

as I don't know how to prove it theoretically .

Pliz help me and tell me,

thanks in advance

Answer 1

You know that $xa+by+zc=1$, now consider $$(ax+by+cz)^{46}=1$$ when expanding this product we get elements on the form $(ax)^i(by)^j(cz)^k$ where $i+j+k=46$. Since $i,j,k\geq 0$, all of the terms will have a factor of either $(ax)^{15}$, $(by)^{16}$, or $(cz)^{17}$. Thus you can write the expression as $$\alpha a^{15}+\beta b^{16}+\gamma c^{17}=1$$

Why $46$? The worst case scenario is when $i=14$, and $j=15$, because then neither $a^{15}$ nor $b^{16}$ is factor of the term, and we want this product then to have a factor of $c^{17}$, that is $j=17$. So $i+j+k=14+15+17=46$.

Answer 2

Hint:

Expand $(xa+yb+zc)^{14+15+16+1}=1$ by the multinomial formula: it is the sum $$\sum_{r+s+t=46}\frac{46!}{r!\,s!\,t!}\,x^ra^r\,y^sb^s\,z^tc^t. $$ and prove that in each term, at least one of the inequalities $$ r\ge 15,\quad s\ge 16,\quad t\ge 17, $$ is true, and group terms for which you can factor out $x^{15}$, then among the remaining terms, group those for which you can factor out $y^{16}$. The other terms will be factorable by $z^{17}$.