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We have the following result:

If $a$, $b$ and $c$ are pairwise coprime integers, such that

$$\frac 1a+\frac 1b=\frac 1c$$

then $a+b$, $a-c$ and $b-c$ are perfect squares.

What I did.

I tried to prove first that $a+b$ is a perfect square.

Multiplying by $abc$ we get

$$c(a+b)=ab.$$

Since $a$ and $c$ are coprime, we get by Gauss' lemma that

$$a\mid a+b.$$

Symmetrically,

$$b\mid a+b.$$

Since $a$ and $b$ are coprime,

$$ab\mid a+b.$$

But we obviously have $a+b\mid ab$, so

$$a+b=ab$$

and $c=1$.

I think this would imply $a=b=2$ which is absurd since $a$ and $b$ are coprime...

The question. Is this going somewhere? What else could I do?

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E. Joseph
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1 Answers1

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The equation $\frac1a+ \frac1b = \frac1c$ is equivalent to $ab=c(a+b) \implies (a−c)(b−c)=c^2$.

Since $a,b,c$ are positive integers, ${1\over{a}}<{1\over{c}}$ and ${1\over{b}}<{1\over{c}}$. Hence $a>c$ and $b>c$. For each pair of integers $x,y $ satisfying $xy=c^2$, we get $a−c=x$ , $b−c=y$.

$\gcd(a,b) = 1$ thus if a prime $p$ divides both $x,y$ we have $p$ dividing both $a,b$; thus $\gcd(x,y) = 1$. And since $xy=c^2$ we have individually $x = z_1^2$ and $y= z_2^2$.

$a−c=x$ , $b−c=y$. Thus $a+b = 2c +x+y = (z_1+z_2)^2$. Hence $a+b$ is a prefect square.

In addition we have also proved that $a-c$ and $b-c$ are perfect squares.

DynamoBlaze
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User8976
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    But if $xy=c^2$, $x=z_1^2$, $y=z_2^2$, then $c=z_1z_1$. But then $a-c=x$ implies $a = c + x = z_1z_2 + z_1^2 = z_1(z_1+z_2)$, and therefore $z_1 \mid \gcd(a,c)=1$ thus $z_1=1$. The same argument for $b-c$ gives $z_2=1$. Thus we get $c=1$, and thus $a=1+1=2$ and $b=1+1=2$, thus $\gcd(a,b)=2\ne 1$. In other words, the conditions cannot be fulfilled. – celtschk Feb 11 '17 at 17:11