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I am attempting to prove that:

$gcd(n^2-100n, 2n+1) = gcd(n-100, 201)$

To do so, I'm making use of the Euclidean algorithm, but I'm not getting to any result that I find particularly useful in demonstrating the equality. When I divide the terms on the LHS, I arrive at a remainder of $\frac{201}{4}$, whereas the RHS yields $\frac{n}{201} - \frac{100}{201}$. I have other incomprehensible calculations that I don't think would add much to the post. Any suggestions would be appreciated, I'm sure the solution is something rather trivial that I'm botching up.

scoopfaze
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  • Note that, as stated in the $6$'th GCD property, "If $m$ is any integer, then $\gcd(a + m\cdot b, b) = \gcd(a, b)$". – John Omielan Apr 08 '22 at 07:20
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    You are doing the right thing (although there are slicker ways to proceed). I too get $\frac{201}{4}$ for the LHS, but your calculation on the RHS is wrong : $201$ goes into $n-100$ $\frac{n}{201}$ times and leaves a remainder of $-100$. So with the convention that GCDs are monic both LHS and RHS evaluate to $1$. – ancient mathematician Apr 08 '22 at 08:33
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    As explained in the dupe such gcds can be evaluated mechanically by a twist on the Euclidean algorithm

    $$\begin{align} (2n!+!1,n^2-100n) &= (2n!+!1,,\color{#c00}2^2(n^2-100n))\ \ \ {\rm by}\ \ (2n!+!1,\color{#c00}2)=1\ &= (2n!+!1,,(\color{#0a0}{2n})^2-200(\color{#0a0}{2n}))\ &= (2n!+!1,,(\color{#0a0}{-1})^2!-200(\color{#0a0}{-1}))\ \ {\rm by}\ \ \color{#0a0}{2n\equiv -1}\pmod{2n+1}\ &= (2n!+!1,,201) \end{align}$$

    Generally this shows $,(2n!+!1,p(n)) = (2n!+!1, 2^k p(-1/2))$ where $k = \deg p,$ and $,p(x),$ is any polynomial with integer coef's.

     – Bill Dubuque Apr 08 '22 at 08:59
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    Similarly if $,(a,b)=1,$ then $$\bbox[8px,border:1px solid #c00]{(an+b,,p(n)) ,=, (an+b,,a^kp(-b/a))},$$ and we can compute $,a^k p(-b/a),$ by a fraction-free method by scaling by $a^k$ as we did above. – Bill Dubuque Apr 08 '22 at 09:21

1 Answers1

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Hint $\gcd(ab,c) = \gcd(b,c)$ whenever $\gcd(a,c) = 1$. Now note that $n^2-100n = n(n-100)$ and check $\gcd(n,2n+1)$.

Ennar
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