How to prove not an integer? The question I am trying to figure out is 
Let n and m be positive integer and let d = gcd(n,m). Define l = nm/d
Let k > d. Prove that at least one of the following statements must be false.
a) nm/k is not an integer
b) n | (nm/k)
c) m | (nm/k)

I thought about making k be d + 1, but don't know what to do from there.
 A: Assume that (b) and (c) are true
(b) n | nm/k $\Rightarrow$ m/k is an integer say $m_k$ (let [nm/k]/n = $m_k$ = m/k) and immediately you see that  k cannot be irrational.
(c) m | nm/k $\Rightarrow$ n/k is an integer say $n_k$
Now nm/k = k.$n_k m_k$ and if (a) is true then this is not an integer $\Rightarrow$ k is not an integer: let k = p/q where q > 1, and p, q are relatively prime. 
Since k = p/q (with q>1) > d then p > d.
From (b) $m_k$ = m/k = m.q/p and as p, q relatively prime then p must divide m. Similarly from (c) p must divide n. So p > d divides both m and n which contradicts d = gcd(m, n).
So, at least one of (a), (b), (c) must be false.
It appears though that (a) is redundant and one of (b) or (c) must be false. The proof is similar. From (b) and (c), $m_k$ = m/k and $n_k$ = n/k. Again, k cannot be irrational so let k = p/q with p, q relatively prime but now allow q $\ge$ 1 so that k could be an integer. Again if k > d then p > d.
$m_k$ = m/k = m.q/p and as p, q relatively prime then p must divide m. Similarly p must divide n. So p > d divides both m and n which contradicts d = gcd(m, n).
