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I start with knowing that two numbers are coprime if:

$n*k + m*j = 1$

So, setting $k = a$ and j = $a+1$ I can solve as follows:

$n*a + m*a + a$

Then,

$a(n+m) + m = 1$

Where can I go from here?

Vaughan Hilts
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    You just need to find a linear combination of $a$ and $a+1$ that give $1$. That is the definition above that you stated, essentially. –  Dec 10 '13 at 20:39
  • @JasonPolak If it's not much of a hassle, could you try and explain a bit more? I'm not quite sure I understand what you mean by "linear combination"? – Vaughan Hilts Dec 10 '13 at 20:48
  • If $a$ and $b$ are any integers, a linear combination (or more precisely, $\mathbb{Z}$-linear combination) of $a$ and $b$ in this setting means an expression of the form $ra + sb$ where $r$ and $s$ are integers. So $4a + 5b$ is an example of an (integer) linear combination of $a$ and $b$. If you can find $r$ and $s$ integers such that $ra + sb = 1$ then $a$ and $b$ are coprime. So you want to find $r$ and $s$, in your case, such that $r(a + 1) + sa = 1$. –  Dec 10 '13 at 21:06

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If not, $p\mid a$ and $p\mid a+1$ would give $p\mid (a+1)-a=1$ for a prime $p$, a contradiction. Or with your definition $1\cdot(a+1)-1\cdot (a)=1$.

Dietrich Burde
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