Prove that $a+b$ is a perfect square

Question

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

Answer 1

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

Answer 2

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 $$ a+b=d^2. $$

Answer 3

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 $k$.