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4 years ago

mycroft holmes
272 Points
```							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
```
4 years ago
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• Test paper with Video Solution
• Mind Map
• Study Planner
• NCERT Solutions
• Discussion Forum
• Previous Year Exam Questions