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 Consider The Factors, Get Perfect Square? (Posted on 2007-09-23)
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.

 See The Solution Submitted by K Sengupta Rating: 3.0000 (3 votes)

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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

 Posted by Brian Smith on 2007-09-23 23:49:37

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