Prove that if p, q, r and s are odd integers, then the equation x^10 + px^9 - qx^7 + rx^4 - s = 0 has no integer roots.
To prove that the equation x^10 + px^9 - qx^7 + rx^4 - s = 0 has no integer roots when p, q, r, and s are odd integers, we can use the technique of contradiction.
First, let's assume that there exists an integer root, let's say x = n, where n is an integer.
Substituting x = n into the equation, we get:
n^10 + pn^9 - qn^7 + rn^4 - s = 0
Now, let's analyze the separate terms in this equation:
1. n^10: The power of n is even, so n^10 will always be a positive integer.
2. pn^9: Since p is an odd integer, pn^9 will have the same parity as n^9, meaning it will also be odd.
3. -qn^7: Similarly, since q is an odd integer, qn^7 will have the same parity as n^7, making it odd as well.
4. rn^4: Since r is odd, rn^4 will have the same parity as n^4, so it can be either even or odd depending on the value of n.
5. -s: Since s is odd, -s will be an odd number.
Now, let's consider the sum of these terms:
n^10 + pn^9 - qn^7 + rn^4 - s = 0
We can see that this sum will always be an odd number, given the parities of each term. However, for the left-hand side to equal zero, the sum must be an even number. This is a contradiction because we have assumed that there exists an integer value, n, that satisfies the equation.
Therefore, our initial assumption is false, and the equation x^10 + px^9 - qx^7 + rx^4 - s = 0 has no integer roots when p, q, r, and s are odd integers.