Proving $f(x)=0$ Has No Integer Solution with Integer Coefficients

  • Context: MHB 
  • Thread starter Thread starter juantheron
  • Start date Start date
  • Tags Tags
    Coefficients Integer
Click For Summary
SUMMARY

The polynomial \(f(x)\) with integer coefficients has been proven to have no integer solutions for \(f(x) = 0\) when both \(f(0)\) and \(f(1)\) are odd numbers. The odd nature of \(f(0)\) indicates that the constant term is odd, while the odd nature of \(f(1)\) necessitates that there is an even number of odd coefficients among the non-constant terms. Consequently, for any integer \(x\), \(f(x)\) remains odd, confirming that \(f(x)\) cannot equal zero.

PREREQUISITES
  • Understanding of polynomial functions and their properties
  • Knowledge of integer coefficients in polynomials
  • Familiarity with parity (odd and even numbers)
  • Basic concepts of number theory
NEXT STEPS
  • Study the implications of polynomial roots in number theory
  • Explore the concept of parity in polynomial equations
  • Learn about the Rational Root Theorem and its applications
  • Investigate further examples of polynomials with integer coefficients
USEFUL FOR

Mathematicians, students studying algebra, and anyone interested in number theory and polynomial equations.

juantheron
Messages
243
Reaction score
1
A polynomial $f(x)$ has Integer Coefficients such that $f(0)$ and $f(1)$ are both odd numbers. prove that $f(x) = 0$ has no Integer solution
 
Mathematics news on Phys.org
jacks said:
A polynomial $f(x)$ has Integer Coefficients such that $f(0)$ and $f(1)$ are both odd numbers. prove that $f(x) = 0$ has no Integer solution


There are details you may need to fill in yourself but:

\(f(0)\) odd implies that the constant term is odd

then \(f(1)\) odd implies that there are an even number of odd coeficients of the non constant terms.

So if \(x \in \mathbb{Z}\) then \(f(x)\) is odd and so cannot be a root of \(f(x)\) CB
 

Similar threads

MHB Polynomial with integer coefficients
  • · Replies 1 ·
Replies
1
Views
2K
MHB Proof: polynomial with integer solutions
  • · Replies 8 ·
Replies
8
Views
3K
MHB Polynomial challenge: Show that not all the coefficients of f(x) are integers.
  • · Replies 3 ·
Replies
3
Views
2K
MHB Compute $\sum_{n=1}^{2020} (f(n)-g(n))$
  • · Replies 1 ·
Replies
1
Views
1K
Graduate Prime Number Powers of Integers and Fermat's Last Theorem
  • · Replies 1 ·
Replies
1
Views
2K
MHB Find Polynomials Fulfilling Real Coefficient Equation
  • · Replies 1 ·
Replies
1
Views
1K
MHB Positive Integer Solutions of $(x^2+y^2)^n=(xy)^{2014}$
  • · Replies 2 ·
Replies
2
Views
2K
MHB Prove that the sum of 6 positive integers is a composite number
  • · Replies 1 ·
Replies
1
Views
1K
Graduate Is this Recursive Pattern in the Collatz Sequence Known?
  • · Replies 3 ·
Replies
3
Views
905
High School Have I proved some part of Fermat's last theorem?
  • · Replies 15 ·
Replies
15
Views
2K