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$a, b, c$ are natural numbers such that $1/a + 1/b = 1/c$ and $gcd(a,b,c)=1$. Prove $a+b$ is a perfect square.

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marked as duplicate by Hakim, Najib Idrissi, user88595, Davide Giraudo, Daniel Fischer Jun 10 '14 at 10:03

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$\dfrac{a+b}{ab} = \dfrac{1}{c} \to a+b = \dfrac{ab}{c}$. Thus:

$c|ab$. Write: $a = dp$, and $b = dq$ with $(p,q) = 1$. Thus: $\dfrac{ab}{c} = d^2\cdot \dfrac{pq}{c}$.

Claim: $c = pq$.

Proof: We have: $b-c = \dfrac{bc}{a}$. Thus: $a|bc \to dp|dqc \to p|qc$. Similarly:

$a - c = \dfrac{ac}{b}$. Thus: $b|ac \to dq|dpc \to q|pc$. Since: $(p,q) = 1$, we have:

$p|c$, and $q|c$. Thus: $pq|c$. So we can write: $c = kpq$. To finish the proof we show: $k = 1$. If $k > 1$, then let $m$ be a prime divisor of $k$, then: from $\dfrac{d^2pq}{c} = \dfrac{d^2pq}{kpq} = \dfrac{d^2}{k} \in \mathbb{N}$, we have: $k|d^2$. So: $m|d^2$ since $m|k$. But $m$ is a prime number, so $m|d$. So: $m|a$, $m|b$, and $m|c$, and $m > 1$. So: $(a,b,c) \geq m > 1$, contradiction. Thus: $k = 1$, and $a+b = d^2$ which is a perfect square.

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Let's $$ c=p_1^{ \alpha _1}\cdots p_k^{\alpha _k}. $$ Since $$ ab=c(a+b), $$ we have $$ a=a_1 p_1^{ \alpha _1}\cdots p_t^{\alpha _t}\\ b=b_1 p_{t+1}^{ \alpha _{t+1}}\cdots p_k^{\alpha _k} $$ for some $1 \leq t \leq k$. It's easy to see that $a_1=b_1=d$, so $$ a+b=d^2. $$

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Why square? May not be.

The formula in General there: Number of solution for $xy +yz + zx = N$

As for the answers solutions: $ab-ac-bc=0$

You can record expanding the square on multipliers: $p^2=ks$

$a=p$

$b=s-p$

$c=p-k$

That the numbers were positive it is necessary that the $s$ had more $k$.

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  • $\begingroup$ It's not clear what you're trying to say. Are you saying the claim that $a+b$ must be a square is wrong? It's not clear how the linked post is supposed to be related to anything here, or even what the linked post is supposed to be saying in the context of the question it was supposed to answer. $\endgroup$ – user2357112 Jun 10 '14 at 9:14

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