We have $\alpha=2n^3-14n+2$ and $\beta=n+3$,
How can we prove that the $\gcd(\alpha,\beta)=\gcd(\beta,10)$,
and what are the possible values for this $\gcd$.
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Actually, i was trying to write $\alpha=\beta*k+10$ – Hedwig Apr 25 '14 at 20:50
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1A quick hint for the second half: if $k$ is a greatest common divisor of $\beta$ and $10$, then it's certainly a divisor of $10$. What are those divisors? – Steven Stadnicki Apr 25 '14 at 21:00
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Hint $\ $ By Euclid, $\,\gcd(n\!-\!a,\,f(n)) = \gcd(n\!-\!a,\,f(n)\ {\rm mod}\ n\!-\!a) = \gcd(n\!-\!a,\color{#c00}{f(a)})\ $ for any polynomial $\,f(n)\,$ with integer coefficients, since $\,{\rm mod} \ n\!-\!a\!:\ n\equiv a\,\Rightarrow\, f(n)\equiv \color{#c00}{f(a)}\ $ by the Polynomial Congruence Rule (or by the Polynomial Remainder Theorem).
Yours is special case $\ f(n) = 2n^3-14n+2,\,\ a=-3,\ $ so $\,\color{#c00}{f(a)} = f(-3) = \color{#c00}{-10}\ $ so the above yields that $\, \gcd(n\!+\!3,f(n)) = \gcd(n\!+\!3,-10) = \gcd(n\!+\!3,10).$
Bill Dubuque
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$2n^3-14n+2 = 2n^2(n+3)-6n(n+3)+4(n+3)-10$ so it easily follows that $\gcd(\alpha, \beta) = \gcd(\beta, -10) = \gcd(\beta, 10)$.
The value of this gcd is $d$ such that $d|10$.
Sandeep Silwal
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