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I cannot tackle the problem.

Let $R$ be a commutative ring with unity. Let $a$, $b$, $c\in R$ be such that there exists $x$, $y$, $z\in R$ with $$xa+yb+zc=1.$$ Show that there exist $\alpha$, $\beta$, $\gamma\in R$ such that $$\alpha a^{15} +\beta b^{16}+\gamma c^{17}=1.$$

Can anyone give a hint how to start? Actually I was thinking about binomial expansion. Will it be helpful or any other way would be better?

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  • $\begingroup$ Seems like a cool question. However you should write it in LaTeX code in the actual question to make it more convenient for everyone. Also you will get community disapproval because your question does describe what you've tried so far. Personally I don't agree with the validity of this disapproval, but in any even, you'll still get it. Feel free to include a description of what you've done so far if you care to avoid it. $\endgroup$ – goblin Apr 3 '18 at 4:23
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    $\begingroup$ This website is a great place to learn basic LaTeX. Just look up an online getting started guide and you'll be right. $\endgroup$ – goblin Apr 3 '18 at 4:41
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    $\begingroup$ Ok i will learn it surely $\endgroup$ – Happy with mathematics Apr 3 '18 at 4:42
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    $\begingroup$ Hint: If you raise the given equation to fourth power, and expand the left hand side, then each term will have at least one of $a^2$, $b^2$, $c^2$ as a factor for otherwise the total exponent $i+j+k$ of $a^ib^jc^k$ would be at most $1+1+1=3<4$. Binomial expansion? Almost! You need trinomials here (but it amounts to much the same thing). $\endgroup$ – Jyrki Lahtonen Apr 3 '18 at 5:22
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    $\begingroup$ There is a corresponding fact involving coprime ideals. But, IIRC, in the texts I've seen the analogue of this result is proven at the level of elements exactly the way I hinted at. Mind you, I'm not very knowledgeable about commutative algebra so others may be able to suggest alternative routes. $\endgroup$ – Jyrki Lahtonen Apr 3 '18 at 6:15

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