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Are $$(a, b, c) = (| 1 |, | 2 |, | 2 |), (| 2 |, | 4 |, | 4 |), (| 2 |, | 3|, | 6 |), (| 1 |, | 1 |, | 1 |)$$ the only integers such that $$\frac{1}{a} + \frac{1}{b} + \frac{1}{c} $$ is an integer?

Xander Henderson
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  • It might help, if you write it as a deviation from a possible solution $$ { 1\over a'+3 } + { 1\over b'+3 }+{ 1\over c'+3 } \overset?\ge 1 $$ and look for nonzero $a',b',c'$. But is this really a question which must appear on the frontside of a math-research board? – Gottfried Helms Nov 12 '16 at 11:34
  • http://math.stackexchange.com/questions/450280/erd%C5%91s-straus-conjecture/831870#831870 – individ Nov 12 '16 at 12:00
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    What about the triple $(3/3/3)$ ? Moreover, there are triples with negative entries doing the job. – Peter Nov 12 '16 at 12:01
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    Note that these triples have a geometric interpretation: the triple $(a, b, c)$ corresponds to a triangle with (interior) angles $(\frac\pi a, \frac\pi b, \frac\pi c)$. The condition $\frac1a+\frac1b+\frac1c=1$ then says that the angles actually form a triangle (since $\frac\pi a+\frac\pi b+\frac\pi c=\pi$), and the condition that $a, b, c$ are all integers mean that an integer number of these triangles will fit around a vertex - in other words, they tile the plane. – Steven Stadnicki Nov 12 '16 at 18:05
  • Does this answer your question? Find all positive integers $(x,y,z)$ such that $\frac{1}{x}+\frac{1}{y}+\frac{1}{z}$ is integer – Matt Hughes Oct 13 '23 at 15:43

2 Answers2

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Assuming we only care about positive integers for the time being, notice that if $\frac{1}{a} + \frac{1}{b} + \frac{1}{c}$ is an integer then it must be at least $1$. If three numbers add up to at least $1$, then at least one of them must be at least $\frac{1}{3}$ - so at least one of $a$, $b$, and $c$ must be no more than $3$.

If none of them is $2$ or less, then $\frac{1}{a}$, $\frac{1}{b}$, and $\frac{1}{c}$ are all no more than a third. But if they add up to at least $1$, then they must all then be exactly $\frac{1}{3}$. So $a = b = c = 3$.

Say the smallest number we have is $2$. Then we can't have two of them (because if $\frac{1}{2} + \frac{1}{2} + \frac{1}{c}$ is a whole number, then $c = 1$ and that's smaller than $2$). But $\frac{1}{2} + \frac{1}{b} + \frac{1}{c}$ can only be at least $1$ if $\frac{1}{b} + \frac{1}{c}$ is at least $\frac{1}{2}$, so we need $\frac{1}{b},\frac{1}{c}$ to be at least $\frac{1}{4}$. Therefore at least one of $b$ and $c$ is at most $4$. So we have either $\frac{1}{2} + \frac{1}{3} + \frac{1}{c}$ (in which case $c = 6$) or $\frac{1}{2} + \frac{1}{4} + \frac{1}{c}$ (in which case $c = 4$).

So far we have $(3, 3, 3)$, $(2, 3, 6)$, and $(2, 4, 4)$, and we've found all of the ones that don't involve a $1$. I'll leave it to you to try to apply this approach to the case when we do have a $1$.

Reese Johnston
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We can consider the positive case, i.e. $a,b,c>0$.

We can categorize into the 3 cases(from simple to copmlex):

1): $a=b=c$. Then $\frac1a+\frac1b+\frac1c=\frac3a\in\mathbb Z$, so $a|3$, hence $a=1,3$.

2): $0<a<b<c$. Then $1\leq\frac1a+\frac1b+\frac1c<\frac3a$, $1\leq a<3$, So $a=1,2$.

If $a=1$, then $\frac1b+\frac1c<\frac22=1$, impossible, so $a=2.$

Then $b\geq3$(since $b>a$), and $$1\leq\frac1a+\frac1b+\frac1c<\frac12+\frac2b\leq\frac12+\frac23<2,$$ thus the sum equals to $1$. Moreover, if $b\geq4$, then $\frac1b+\frac1c<\frac24$, which means the sum is not an integer. So $a=2,~b=3~,\Rightarrow c=6.$

3): $a=b<c$. By the method of 2), we have $1\leq\frac2a+\frac1c<\frac3a,~\Rightarrow a<3$.

If $a=1$, then $\frac1c<1$, impossible.

If $a=2$, $\frac1c<1$, still impossible.

Hence, in this case, there is no result.

4): $a=b>c$. By the above argument, $1\leq\frac2a+\frac1c<\frac3c$. $c=1,2$.

If $c=1$, then $a=2$.

If $c=2$, $\frac2a\in \mathbb Z+\frac12$, and $a\geq3$. So $a=4$.

PS: For the non-positive case, there are infinite solutions.

e.g. $a=b=2n,~c=-n$, where $n\in \mathbb N^+.$

DLIN
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