Proving a polynomial has no integer solution

Thread starter timetraveller123
Start date Jun 6, 2017
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Integer Polynomial

In summary: Sorry, I misread what you wrote. You're saying that if ∣r2r3...rn∣=12∣r2r3...rn∣=12\mid r_2 r_3 ...r_n\mid=\frac 12 then (∣an∣)=2(∣an∣)=2(\mid a_n \mid)=2. Is that what you meant to say?

Jun 8, 2017

timetraveller123

oh now i see i am too used to seeing polynomials in the standard form thanks a lot to everyone
@haruspex @Buffu @Dick

<h2>1. What does it mean for a polynomial to have no integer solution?</h2><p>When a polynomial has no integer solution, it means that there is no value for the variable in the polynomial that would result in a whole number when plugged in. In other words, there are no integers that satisfy the equation.</p><h2>2. How can I prove that a polynomial has no integer solution?</h2><p>There are a few methods for proving that a polynomial has no integer solution. One way is to use the Rational Root Theorem to test all possible rational roots of the polynomial. If none of these roots result in a whole number, then the polynomial has no integer solution. Another method is to use the Fundamental Theorem of Algebra, which states that a polynomial of degree n has at most n complex roots. If the polynomial has a degree of n and no integer roots, then it must have no roots at all.</p><h2>3. Can a polynomial have no integer solution but still have real solutions?</h2><p>Yes, it is possible for a polynomial to have no integer solution but still have real solutions. For example, the polynomial x^2 + 1 has no integer solution, but it has two complex solutions (i.e. solutions that involve the imaginary number i).</p><h2>4. Are there any shortcuts for proving that a polynomial has no integer solution?</h2><p>Unfortunately, there are no shortcuts for proving that a polynomial has no integer solution. The best way to prove this is to use the methods mentioned in question 2, such as the Rational Root Theorem or the Fundamental Theorem of Algebra.</p><h2>5. Can a polynomial have an infinite number of integer solutions?</h2><p>No, a polynomial cannot have an infinite number of integer solutions. This is because a polynomial of degree n can have at most n distinct roots. Therefore, the number of integer solutions is limited by the degree of the polynomial.</p>

1. What does it mean for a polynomial to have no integer solution?

When a polynomial has no integer solution, it means that there is no value for the variable in the polynomial that would result in a whole number when plugged in. In other words, there are no integers that satisfy the equation.

2. How can I prove that a polynomial has no integer solution?

There are a few methods for proving that a polynomial has no integer solution. One way is to use the Rational Root Theorem to test all possible rational roots of the polynomial. If none of these roots result in a whole number, then the polynomial has no integer solution. Another method is to use the Fundamental Theorem of Algebra, which states that a polynomial of degree n has at most n complex roots. If the polynomial has a degree of n and no integer roots, then it must have no roots at all.

3. Can a polynomial have no integer solution but still have real solutions?

Yes, it is possible for a polynomial to have no integer solution but still have real solutions. For example, the polynomial x^2 + 1 has no integer solution, but it has two complex solutions (i.e. solutions that involve the imaginary number i).

4. Are there any shortcuts for proving that a polynomial has no integer solution?

Unfortunately, there are no shortcuts for proving that a polynomial has no integer solution. The best way to prove this is to use the methods mentioned in question 2, such as the Rational Root Theorem or the Fundamental Theorem of Algebra.

5. Can a polynomial have an infinite number of integer solutions?

No, a polynomial cannot have an infinite number of integer solutions. This is because a polynomial of degree n can have at most n distinct roots. Therefore, the number of integer solutions is limited by the degree of the polynomial.