Let $k$ be a positive integer. Prove that for any integer $n>=k+1$, if $k|n$, then $gcd(n,n-k) = k$.
My attempt:
$q_{1}n + q_{2}(n-k) = k$
Also, $k|n = k*u = n$
$q_{1}*k*u + q_{2}(k*u-k) = k$
I don't know what do do after this step
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Are you familiar with the Euclidean algorithm? If so, I suggest you use it. – Arthur Apr 19 '17 at 22:25
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If $n=km$, then $n-k=k(m-1)$. Reduce this to finding the GCD of $m$ and $m-1$, and you can finish it – Clayton Apr 19 '17 at 22:27
2 Answers
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You tell nothing about $q_1$ and $q_2$, so this path leads to nothing.
What you have to show is that $k$ is the maximum common divisor.
The hypothesis tells you $n=mk$, so also $n-k=(m-1)k$. Therefore $k$ is a common divisor of $n$ and $n-k$.
If $h$ is a common divisor of $n$ and $n-k$, then $$ n=ah,\qquad n-k=bh $$ and therefore $$ k=n-(n-k)=ah-bh=(a-b)h $$ Hence $h\le k$ (actually $h$ is a divisor of $k$).
egreg
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1@Ag_Yog $\gcd(n,n-k)$ is what you have to prove! If $n=mk$, then $n-k=mk-k=(m-1)k$. – egreg Apr 19 '17 at 23:02
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Notice that since $k|n$, we have $\gcd(n,k)=\gcd(n,-k)=k$. This is because $k|k$ and $k|n$, and there are no greater divisors of $k$.
Further, it is not to hard to see that $\gcd(a,b)=\gcd(a,a+b)$. (See for instance, Prove: $\gcd(a,b) = \gcd(a, b + at)$.)
Thus, we have $k=\gcd(n,-k)=\gcd(n,n-k)$, where we used the above fact.
helloworld112358
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