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.
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Submitted by K Sengupta
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Rating: 3.0000 (3 votes)
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Solution:
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(Hide)
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1/p + 1/q = 1/r
or, (p-r)(q-r) = r^2……(#)
Let s be any prime such that s divides r. Then, from (#) it follows that s divides (p-r)(q-r). If possible, let s divide both (p-r) and (q-r). then, it follows that s must divide both p and q in addition to r. This is a contradiction since no integer >=2 can divide p, q and r simultaneously.
Accordingly, either s can divide (p-r) , or s can divide (q-r). Thus, there do not exist any integer >=2 dividing both (p-r) and (q-r). Therefore, gcd(p-r, q-r) = 1. But from (#), (p-r)(q-r) is a perfect square. This is only possible iff (p-r) and (q-r) separately correspond to perfect squares.
Hence,
p-r = t^2 (say), so that:
q-r = r^2/t^2
Then, p+q – 2r = t^2 + r^2/t^2
Or, p+q = t^2 + r^2/t^2 + 2r
Or, p+q = (t + r/t)2
Consequently, (p+q) is a perfect square.
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For alternative methodologies refer to the solution posted by Chesca Ciprian in this location, and by Brian Smith in this location .
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