Hello I have to determine the gcd of $n^3+3n^2-5 $ and $ n+2$, but I can’t do it because I know we have to eliminate the $n$. I tried to eliminate $n^3 $ but I’m stuck for the rest.
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You are allowed to multiply either one by an expression including $n$, then subtract. That should allow you to clear the $n^3$ if you choose the expression well. – Ross Millikan May 15 '21 at 14:00
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$$n^3+3n^2-5-n^2(n+2)=n^2-5$$ – lone student May 15 '21 at 14:01
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do you know that $gcd(a,b) = gcd(a,b \pm k)$ where $k \in Z$ – Aditya_math May 15 '21 at 14:03
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Follows immediately from the Euclidean Algorithm or Remainder Theorem as explained in the linked dupes, i.e. $,\gcd(f(n),n+2) = \gcd(f(-2),n+2),$ for any polynomial $,f(x),$ with integer coefficients. – Bill Dubuque May 15 '21 at 20:24
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$$n^3 + 3n^2 - 5 - n^2 (n+2) = n^2 - 5$$ and $$n^2 - 5 - n(n+2) = -2n-5$$ Why did I do this? $\gcd(f(x),g(x)) = \gcd(f(x) - \lambda g(x), g(x))$ where $\lambda$ is an integer. So, keeping the above two equations in mind we have $$\gcd(n^3 + 3n^2 - 5, n+2) = \gcd(-2n-5, n+2) = 1$$ i.e. the two are co-prime (the greatest common divisor is $1$).
stoic-santiago
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Let $g$ be their gcd. Then $$n^3 +3n^2 -5 = ga \\ n+2 = gb $$ Subtract the first from the cube of the second: $$3n^2 +12n +13 = g^3b^3 -ga $$ Subtract from this thrice the square of the second equation: $$1 = g^3b^3 -ga-3g^2b^2 = g(g^2 b^3 -a-3gb^2)$$ This implies $$g=1$$
Vishu
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Do you know how to search for dupes, e.g. using https://approach0.xyz/search/ ? – Bill Dubuque May 15 '21 at 20:51
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@BillDubuque Yes, but it’s kind of inconvenient to do this for every question I answer. – Vishu May 15 '21 at 21:31