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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$\leqq b\leqq c$,

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    $\begingroup$ Did you find any such triples? $\endgroup$ – coffeemath Jan 19 at 22:44
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    $\begingroup$ 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. $\endgroup$ – Henning Makholm Jan 19 at 22:51
  • $\begingroup$ It's wolog that $a = \min(a,b,c)$ but it's not wolog that $\mid(a,b,c)|\min(a,b,c)+1$ $\endgroup$ – fleablood Jan 19 at 23:10
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Hint: given $a \le b$ and $\frac {a+1}b$ is an integer, you must have $b=a+1$ or $a=b=1$.

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

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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.

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