# Find all positive triples of positive integers a, b, c so that $\frac {a+1}{b}$ , $\frac {b+1}{c}$, $\frac {c+1}{a}$ are also integers.

Find all positive triples of positive integers a, b, c so that $$\frac {a+1}{b}$$ , $$\frac {b+1}{c}$$, $$\frac {c+1}{a}$$ are also integers.

WLOG, let a$$\leqq b\leqq c$$,

• Did you find any such triples? – coffeemath Jan 19 at 22:44
• Are you sure that is WLOG? The condition is not invariant under arbitrary permutations of $a,b,c$ but only under 3-cycles. A priori, there might be a solution where, as you go around the cycle, there are two decreasing and one increasing step. – Henning Makholm Jan 19 at 22:51
• It's wolog that $a = \min(a,b,c)$ but it's not wolog that $\mid(a,b,c)|\min(a,b,c)+1$ – fleablood Jan 19 at 23:10

Hint: given $$a \le b$$ and $$\frac {a+1}b$$ is an integer, you must have $$b=a+1$$ or $$a=b=1$$.

• Or $a = b = 1$. – Henning Makholm Jan 19 at 22:49
• @HenningMakholm: Good point. Thanks. – Ross Millikan Jan 19 at 22:51

If any two of $$a,b,c$$ are equal, then wlog. $$a=b$$. As $$\frac{b+1}{a}=1+\frac1a$$ is an integer, we conclude $$a=b=1$$. The remaining conditions are that $$\frac{c+1}{1}$$ and $$\frac 2c$$ are integers, which lead us to the solutions $$(1,1,1),\qquad (1,1,2)$$ (and cyclic permutations of the latter).

So assume $$a,b,c$$ are pairwise different. By cyclic permutation, we may assume wlog that $$a or that $$a>b>c$$. In the first case, $$0<\frac{a+1}{b}\le \frac bb=1$$ and hence $$a+1=b$$. Likewise, $$b+1=c$$. Then the last integer is $$\frac{c+1}a=\frac{a+3}a=1+\frac 3a$$ and we must have $$a=1$$ or $$a=3$$, whic gives us the solutions $$(1,2,3),\qquad (3,4,5)$$ (and cyclic permutations).

In the case $$a>b>c$$, we instead have that $$0<\frac{c+1}{a}\le \frac{c+1}{c+2}<1$$, not an integer. So this case does not produce additional solutions.

If $$a \le b$$ but $$b|a+1$$ then $$b \le a+1$$ so either $$a = b$$ or $$b=a+1$$.

If $$a = b$$ then $$b|b+1$$ and $$a=b=1$$. The you $$c|b+1=2$$ and $$b|c+1$$. So either $$c=1$$ or $$c = 2$$.

So so far we have $$(1,1,1)$$ or $$(1,1,2)$$

If $$b = a+1$$ then $$c|a+2$$ so $$c \le a+2$$ but $$a < b=a+1 \le c\le a+2$$ so either $$c = b = a+1$$ or $$c= a+2$$.

If $$c= b =a+1$$ then we have $$b|a+1 = b$$ and $$c=b|b+1$$ and $$a=b-1|c+1=b+1$$. So $$b=1$$ but then $$a=0$$ and that's a contradiction.

If $$c=a+2$$ and $$b= a+1$$ we have: $$b|a+1=b$$; $$c|b+1 =c$$ and $$a|c+1 = a+3$$. This means $$a|3$$ and $$a =1$$ or $$a =3$$.

So we have $$(1,2,3)$$ or $$(3,4,5)$$.

To double check:

$$(a,b,c) = (1,1,1)\implies$$ $$b|a+1, c|b+1, a|c+1$$ all translate to $$1|2$$. which is true

$$(a,b,c) = (1,1,2) \implies b=1|a+1=2; c=2|b+1=2; a=1|c+1=3$$. All true.

$$(a,b,c) = (1,2,3)\implies b=2|a+1=2; c=3|b+1=3; a=1|c+1 = 4$$. All true.

$$(a,b,c) = (3,4,5)\implies b=4|a+1=4; c=5|b+1=5; a=3|c+1=6$$. All true.