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A|B, A|C A|B±C

A|N A ≤ N

3N+5|3(7N+1) → 3N+5|21N+3

3N+5|7(3N+5)→ 3N+5|21N+35

3N+5|(21N+35)-(21N+3) → 3N+5|32

3N+5 ≤ 32

3N ≤ 27

N ≤ 9

N=1 → 3(1)+5|7(1)+1 = 8|8 = 1

N=9 → 3(9)+5|7(9)+1 = 32|64 = 2

If $3n+5|7n+1$ , then there exists a positive integer $k$ such that $7n+1=k(3n+5)$ $7n+1=3kn+5k$ $7n-3kn=5k-1$ $n(7-3k)=5k-1$ $n=\frac{5k-1}{7-3k}$

As $n>0$ , we have $\frac{5k-1}{7-3k}>0$ $\frac{1}{5}<k<\frac{7}{3}$

The only positive integer solutions for $k$ above is $k=1,2$ . Test these out and we find two distinct integer solutions for $n$ .

Took 15 seconds to solve.