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Grade:12

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mycroft holmes
272 Points
7 years ago
Presenting a proof I read on math[dot]stackexchange
 
We have f(x) = P^2(x)+Q^2(x) and f(0) = 1000 which is a multiple of 4.
 
That means P(0) and Q(0) are both even (else P2(0)+Q2(0) would not be of the form 4n). That means the constant term of P(x) and Q(x) is even. Now its easily seen that this means that the coeff of x in both P2(x) and Q2(x) is a multiple of 4.
 
Now if g(x) = f(x)+2x = R2(x)+S2(x) with g(0) = 1000, then we have the coeff ox on LHS is of the form 4n+2, whereas coeff of x on RHS is of the form 4n which is a contradiction.
 
So, f(x)+2x will not be a Fermar Polynomial

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