p, q and r are

*three positive integers* satisfying 1/p + 1/q = 1/r, such that there is no integer ≥ 2 dividing p, q and r simultaneously. However, p, q and r are not necessarily pairwise

coprime .

*Prove that (p+q) is a perfect square*.

From "Reciprocal Equation #2" parameterize r,p,q as:

r = a

p = a+x

q = a + a^2/x

where x is a factor of a^2

Let f = gcd(A,x)

Then there exist B and y such that A=f*B, x=f*y, and gcd(B,y)=1

Then:

r = f*B

p = f*B+f*y

q = f*B + (f*B^2)/y

where y is a factor of f*B^2

With gcd(B,y)=1 that forces y to be factor of f, but in order to satisfy gcd(r,p,q) = 1, y must be a multiple of f. Therefore y=f.

With that:

r = y*B

p = y*B + y^2

q = y*B + B^2

where gcd(B,y) = 1

Then p+q = (y*B + y^2) + (y*B + B^2) = (y + B)^2