Prove that the Diophantine equation
$\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=1$
has only one solution for positive integers $a,b,c$ with $a<b<c$
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@DietrichBurde yes but the question is to prove that there exist only one – Satvik Mashkaria Jan 07 '15 at 16:19
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I saw this on MSE already, but can't find the duplicate now. But see here for ideas. – Dietrich Burde Jan 07 '15 at 16:26
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The formula in General there. http://math.stackexchange.com/questions/450280/erdös-straus-conjecture/831830#831830 – individ Jan 07 '15 at 17:06
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there must be a number smaller than $3$ since there is going to be a number larger than $3$. So $a=2$. we now want $\frac{1}{b}+\frac{1}{c}=\frac{1}{2}$ with $b$ and $c$ larger than $2$. If we take $\frac{1}{4}$ we need to repeat it, so we need a number larger than $\frac{1}{4}$, this forces us to take $\frac{1}{3}$.
Asinomás
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1Probably easier to say if the average of three distinct numbers is $\frac 1{3}$, then one of them is bigger than $\frac{1}3$, rather than the more peculiar "there must be a number smaller, so there must be a number larger" argument. Same for the $\frac 12$ case - the average of two distinct numbers is $\frac 14$, so one of them must be larger than $\frac 14$. – Thomas Andrews Jan 07 '15 at 16:40
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Hint:
$$1\geq\frac{1}{a}>\frac{1}{b}>\frac{1}{c}$$ so , $$\frac{1}{a}+\frac{1}{b}+\frac{1}{c}<\frac{3}{a}$$.
If $a>3$ the sum is less than $1$.
Check the cases $a=2$ , $a=1$.
Konstantinos Gaitanas
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