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juantheron
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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
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
To prove that a function has no integer solution, you must first analyze the function and its coefficients. Then, you can use various methods such as the Rational Root Theorem, substitution, and contradiction to show that there are no possible integer values that satisfy the equation.
Some common techniques used to prove that $f(x)=0$ has no integer solution include the Rational Root Theorem, substitution, and contradiction. Other methods such as the Remainder Theorem, Euclidean Algorithm, and modular arithmetic can also be useful.
Yes, a function can have no integer solution but still have real solutions. For example, the function $f(x)=x^2+1$ has no integer solutions but has two real solutions, $x=1$ and $x=-1$.
Yes, there are special cases where a function with integer coefficients has no integer solution. One example is when the function has no real solutions, such as $f(x)=x^2+1$, as mentioned earlier. Another example is when the function has complex solutions, such as $f(x)=x^2+4$, which has two complex solutions, $x=2i$ and $x=-2i$.
No, there is no guaranteed method to prove that a function has no integer solution. However, by utilizing various techniques and methods, it is possible to prove that a function has no integer solution with a high degree of confidence. It is important to carefully analyze the function and its coefficients and use multiple approaches to support the proof.