Finding $m+n$ from $m+n+mn+1=91$

Question

If $m,n$ are natural numbers such that $m+n+mn+1=91$ .Then how to find $m+n$

Are you aware that there's actually multiple solutions to the equation and hence multiple values $m+n$ can be? — Noble., Feb 27 '13 at 19:27
@Noble First half is true, second half is false (unless you include $0$ in the natural numbers). — Erick Wong, Feb 27 '13 at 19:28
@ErickWong Ah right, I was actually including $0 \in \mathbb{N}$ which is the convention my course takes. — Noble., Feb 27 '13 at 19:37
@DonAntonio-correct, as for (m+1)(n+1)=mn+m+n+1=91,m+1=7 and n+1=13 or vice versa,because (612)+6+12+1=(126)+12+6+1=91, — , Mar 27 '16 at 11:33

score 4 · Answer 1 · answered Feb 27 '13 at 19:27

4

Hint: $$m+n+mn+1=(m+1)(n+1).$$

answered Feb 27 '13 at 19:27

Cameron Buie

score 2 · Accepted Answer · answered Feb 27 '13 at 19:28

2

$$mn+m+n+1=91\Longrightarrow m(n+1)+(n+1)=91\Longrightarrow (m+1)(n+1)=91$$

But $\,91=7\cdot 13\,$ , so...

answered Feb 27 '13 at 19:28

DonAntonio

2 Answers2