Question

Let \(r\) and \(s\) be two real or complex numbers and m and n be two integers \
such that
\[r + s = m \\ r s = n\] and let \(k\ge 1\) be an integer.Show that
\[r^k + s^k\] is an integer.
You can prove it by the methods used to prove
http : // openstudy.com/study #/updates/5498 ce78e4b0b8a54fae51fb

Answer 1

http://openstudy.com/users/eliassaab#/updates/5498ce78e4b0b8a54fae51fb

Answer 2

strange having to prove this. idk how to prove it.
you have a sum of numbers and both numbers are integers raised to the power of integers, so why wouldn't they be integers?

Answer 3

I know this is not enough, but I was just wondering.

Answer 4

r and s are not necessarily integers.

Answer 5

yes but you can substitute m and n for them, and that will make it then.

Answer 6

I wish I can help, sorry.

Answer 7

Let's start by defining:\[r^k + s^k = (r+s)^k - \sum_{i=1}^{k-1}\left(\begin{matrix}k \\ i\end{matrix}\right)r^is^{k-i}\]Since we know k is an integer greater than 0, this is just a simple reexpression of Pascal's Rule.
By the given parameters of the problem, r+s is an integer, so (r+s)^k is an integer, so we simply have to prove that the value of the summation is an integer. If k is even, then when i = k/2, i=k-i. Since the value of rs is an integer, the value of \[\left(\begin{matrix}k/2 \\ k/2\end{matrix}\right)r^{k/2}*s^{k/2}\] will be an integer.
For each other value j of i such that j < k/2, there will be a related value k-j of i. Then consider the arbitrary case of one such pair of values:
\[\left(\begin{matrix}k \\ j\end{matrix}\right)r^{j}*s^{k-j} + \left(\begin{matrix}k \\ k-j\end{matrix}\right)r^{k-j}*s^{j} =\]\[\left(\begin{matrix}k \\ j\end{matrix}\right)r^j*s^j*s^{k-2j} + \left(\begin{matrix}k \\ j\end{matrix}\right)r^j*s^j*r^{k-2j} =\]\[(\left(\begin{matrix}k \\ j\end{matrix}\right)r^j*s^j)(s^{k-2j} + r^{k-2j})\] Since rs is an integer and r+s is an integer, this pair of terms has an integer value. Since each term of the summation is either a member of such a pair, or has an integer value, the value of the summation is an integer and as such the value of \[r^k + s^k\]is an integer.

Answer 8

Oops, I left out a step. Since j is an integer and j < k/2, k - 2j is an integer greater than 0. Thus, we can use induction to prove that \[s^{k-2j}+r^{k-2j}\] being a lower-exponent case of the same problem, is an integer.

Since s^1 + r^1 = s + r is an integer, we have our base case.

Answer 9

If you look at my post @Blacksteel in http://openstudy.com/users/eliassaab#/updates/5498ce78e4b0b8a54fae51fb
you can write an easier proof.

Answer 10

I'm not clear on why the assumption that r and s are roots of a polynomial equation holds. The original problem you based your solution on explicitly states that, but using a trivial case like r = 2, s = 3, m = 5, n = 6, all of the assumptions of this problem hold and r and s are clearly not roots of \(x^2 + x - 1 = 0\)

Answer 11

No, they are the solutions of a similar but not the same quadratic. But, you are right in the
fact that I should mention r and s could be real or complex number. I have edited the problem to reflect this.

Answer 12

That wasn't my point at all - your solution assumes that r and s are the roots of the SAME polynomial. If they aren't, you can't construct an equation with \(r^{k-2}\) and \(s^{k-2}\) having the same coefficient, which means you can't use induction. I haven't done due diligence to state that your solution isn't applicable with modifications here, but it definitely doesn't translate directly.

Answer 13

Take the quadratic \[ x^2 - m x + n =0\] This equation has two real or complex solutions
r and s, such that r+s=m and r s =n

Answer 14

Do you know the sum and product formula for the two roots of a quadratic?

Answer 15

*

Answer 16

The sum (S) and the product (P) of real or complex roots of
\[
a x^2 + b x + c=0\\ \text { are }\\
S= -\frac b a\\
P = \frac c a
\]

Answer 17

Okay, that's the part I was missing. (I know the sum and product formulas and how to construct a polynomial with given roots, but you didn't state in the original proof that that was what you used and I wasn't sure how you got there, and therefore how to generalize it).