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I'm doing this question for an assignment, the question is:

Prove: if $k\ \epsilon\ \mathbb N$, then $gcd(3k+2, 5k+3)=1$.

I was going to do it by induction, so what I have so far is:

$n\ |\ 3k+2$ implies that $3k+2=ln$ and $n\ |\ 5k+3$ implies that $5k+3=mn$ for some $l,n\ \epsilon\ \mathbb Z$.

Then, for $k=1:gcd(3(1)+2, 5(1)+3)=1$, which is a true statement.

For $k=k+1:gcd(3k+5,5k+8)=1$

I know also that $l(3k+5)+m(5k+8)=1$, but I'm not sure what to do next.

Bill Dubuque
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gabforse
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    This is a problem of divisibility: generally in these statements induction is not a good idea (what have divisors of $n$ and divisors of $n+1$ in common?). – Crostul Mar 09 '16 at 14:13

2 Answers2

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You can directly use Euclid's algorithm, by the way: $$\gcd(3k+2, 5k+3) = \gcd(3k+2, 2k+1) = \gcd(k+1, 2k+1) = \gcd(k+1, k) = 1$$

Patrick Stevens
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How about this: $$5\cdot (3k+2) + (-3)\cdot (5k + 3) = 1.$$

You know that $a$ and $b$ are relatively prime, iff there exist integers $m$ and $n$, such that $ma+nb=1$.

Mankind
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