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**emakarov** OK, here is a sketch. Suppose that p, q, r are prime, i, j₁, j₂ and k are positive integers, and $\displaystyle x=p^iq^{j_1}$, $\displaystyle y=q^{j_2}r^k$. In the general case, the variables here should be vectors. Without loss of generality, let $\displaystyle j_1\le j_2$. Then

$\displaystyle z=\frac{xy}{x+y} =\frac{p^iq^{j_1+j_2}r^k}{q^{j_1}(p^i+q^{j_2-j_1}r^k)} =\frac{p^iq^{j_2}r^k}{p^i+q^{j_2-j_1}r^k}$.

Let $\displaystyle s=p^i+q^{j_2-j_1}r^k}$. Then the only primes that could divide s are p, q and r. If p | s, then $\displaystyle p\mid q^{j_2-j_1}r^k$, a contradiction. Similarly, r | s is false. Thus $\displaystyle s=q^j$ for some j. Moreover, $\displaystyle j=j_2$; otherwise q ∈ GCD(x, y, z). Finally, if $\displaystyle j_2>j_1$, then q | s implies $\displaystyle q\mid p^i$, a contradiction. Therefore, $\displaystyle j_1=j_2=j$, $\displaystyle s=q^j$ and $\displaystyle x+y=q^{2j}$.

See if you can shorten this and if it can be extended to the general case.