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On 29 Apr 2006 19:25:48 -0700, "Deep" <deepkdeb@xxxxxxxxx> wrote:
>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)
You haven't actually proved anything.
Trying your logic, can we then say that the equation x+2=5 has no
solutions because "it's an equation, not an identity"?
Of course not.
If an equation is an identity, then any legal substitution yields a
solution, but an equation which is not an identity may or may not have
solutions, depending on the equation.
Examples:
The equation 2*x = x + x is an identity. Any real number is a
solution.
The equation 2*x = x + 1 is not an identity, but has a solution.
The equation x = x + 1 is not an identity, but has no solutions.
quasi
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