I have a problem with elemental number theory. I started with the expression $$ (a - \frac{1}{b})(b - \frac{1}{c})(c - \frac{1}{a}) $$

and task to find all natural $a,b,c$ so that the result of the expression is an integer.

I managed to show that-

$$ a | bc - 1$$

$$b | ac - 1$$

$$c | ab - 1 $$

But I don't know how to solve this 3 equations. Is there even enough data?

  • $\begingroup$ I think you mean $a\mid bc+1$, etc., not $bc-1$. $\endgroup$ – Barry Cipra Apr 17 '14 at 17:34
  • $\begingroup$ If you multiply the three terms and throw away the obvious integers, what remains? $\endgroup$ – Karolis Juodelė Apr 17 '14 at 17:37
  • $\begingroup$ Sorry, I wrote the original expression with + instead of -. Fixed it already. $\endgroup$ – user143863 Apr 17 '14 at 20:09
  • $\begingroup$ $a=2,b=3,c=5$ and its permutations: $(2-\frac{1}{3})(3-\frac{1}{5})(5-\frac{1}{2})=21$. Also "trivial" solutions $a=b=1, c\in \mathbb{N}$: $(1-1)(1-\frac{1}{c})(c-1)=0$. $\endgroup$ – Oleg567 Apr 18 '14 at 14:09

Expand to get $abc|ab+ac+bc+1$.

Then $\frac{1}{a}+\frac{1}b+\frac{1}c+\frac{1}{abc}$ must be an integer. If $3 \le a \le b \le c$,

then $\frac{1}{a}+\frac{1}b+\frac{1}c+\frac{1}{abc} \le \frac{28}{27}$. Now easy casework. ;)

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  • $\begingroup$ With the corrected basic expression it is even simpler. You get a minus before $$\frac{1}{abc}$$. So if understand it correctly I just have to check the 6 possible combinations and voila. Thanks man :) i feel so dumb having tried to solve this all day long. $\endgroup$ – user143863 Apr 17 '14 at 20:35

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