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Old 2021-06-08, 00:13   #8
paulunderwood
 
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Sep 2002
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What is wrong with this argument?

Let:
x^2-P*x+Q=0
y^2-P*y-Q=0
s^2-P*s+R=0
t^2-P*t-R=0

Then:
x^2+y^2-P*(x+y)=0
s^2+t^2-P*(s+t)=0

Or
P=(x^2+y^2)/(x+y)=(s^2+t^2)/(s+t)

That is since gcd(P,n)=1:
(x^2+y^2)*(s+t)=(s^2+t^2)*(x+y)

And the the following must be true since gcd(x^2+y^2, x+y) is either 1 or 2:
x+y=s+t

And so:
x^2+y^2=(x+y)^2-2*x*y=s^2+t^2=(s+t)^2-2*s*t

So that:x*y=s*t

If a prime p divides x it must divide s or t

Wlog then:
x^2-P*x+Q==x^2-P*x+R

I.e:
p divides Q-R, but a condition is that gcd(Q^2-R^2,n)=1.

QED
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