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Lemma:
The following equation has no solutions under the given conditions.
Qq^1/2 + Rr^1/2 = Ss^1/2 + Tt^1/2 (1)
Conditions: Q, R, S, T are integers > 1; q, r, s, t are square free
integers > 1.
(Qq, Rr) = 1
Proof:
One obtains (2) from (1)
a^1/2 + b^1/2 = c^1/2 + d^1/2 (2)
Where Qq^1/2 = a^1/2, Rr^1/2 = b^1/2, Ss^1/2 = c^1/2 and Tt^1/2 = d^1/2
All of a, b, c, d are integers > 1 and none is a perfect square.
When c is different from a and b (2) has no solutions[1] in integers
If c = a then d = b. In that case (2) and hence (1) becomes an
identity.
This is a contradiction because (1) is an equation not an identity.
This Proves the Lemma.
Any comment upon the correctness of the proof will be appreciated.
[1] L. E. Dickson,History of the theory of numbers,Vol.2 (1920)
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