Question

Let m and n be two integers and p a positive integer, show that
\[\left(m-\sqrt{m^2-4 n}\right)^p+\left(\sqrt{m^2-4 n}+m\right)^p
\]
is an integer divisible by \[ 2^p\]

Answer 1

Every day it seems like you ask a similar question, but they get harder and harder haha. I am going to try to solve it this time.

Answer 2

this question is equivalent to the previous two questions ;p

Answer 3

\[x^2 - mx + n = 0\]
\[x = \frac{m \pm \sqrt{m^2-4n}}{2}\]

prove that \(\alpha^p + \beta^p\) is an integer

Answer 4

Essentially it looks like these are just two roots of a quadratic equation. Naturally solving them will bring out 1/2 and you can simply raise both of these to arbitrary powers of p and the two will come along for the ride.

Answer 5

yes we still need to prove it though
if i remember, the recent proof we had was for a simpler quadratic x^2 + x = 1

Answer 6

I'll try induction on \(p\) :

\(F(p) = \left(m-\sqrt{m^2-4 n}\right)^p+\left(\sqrt{m^2-4 n}+m\right)^p\) is divisible by \(2^p\)

\(F(1) = 2^1m \)
\(F(2) = 2^2(m^2-2n)\)
the statement holds for \(F(1)\) and \(F(2)\)

suppose \(F(k)\) and \(F(k-1)\) are divisible by \(2^k\) and \(2^{k-1}\)
then we have \(F(k) = 2^k x\) and \(F(k-1) = 2^{k-1}y\) for some integers \(x\) and \(y\)

\[\begin{align}F(k+1) &= \left(m-\sqrt{m^2-4 n}\right)^{k+1}+\left(\sqrt{m^2-4 n}+m\right)^{k+1}\\~\\

&= 2m\times F(k) - 4n\times F(k-1)~~~\color{red}{\star}\\~\\
&= 2m\times 2^k x - 4n\times 2^{k-1}y\\~\\
&= 2^{k+1}(mx - ny)
\end{align}\]
\(\blacksquare\)

\(\color{red}{*} : ~ a^{k+1}+b^{k+1} = (a+b)(a^k+b^k) - ab(a^{k-1}+b^{k-1})\)

Answer 7

\[r = \frac {1} {2} \left (m - \sqrt {m^2 - 4 n} \right) \\
s = \frac {1} {2} \left (m + \sqrt {m^2 - 4 n} \right) \\
r + s = m \\
r s = n \\
\]
By the problem I posted before this one,
we have \[ r^p + s^p \] is an integer say q,
\[\left(m-\sqrt{m^2-4 n}\right)^p+\left(\sqrt{m^2-4 n}+m\right)^p=2^p
\left(r^p+s^p\right)==2^p q
\] We are done

Answer 8

@ganeshie8

Answer 9

Ahh yes it nicely packs the previous work xD I see now we may choose arbitrary m, n and a generate whole set of fun divisibility problems!

Answer 10

If I did this correctly, we can show it's also divisible by 2^(2p)

\[ \left( m-\sqrt{m^2-4 n} \right)^p+\left( \sqrt{m^2-4 n}+m \right)^p \\
\frac{\left(m+\sqrt{m^2-4 n}\right)^p}{\left(m+\sqrt{m^2-4 n}\right)^p}\left(m-\sqrt{m^2-4 n}\right)^p+\frac{\left(m-\sqrt{m^2-4 n}\right)^p}{\left(m-\sqrt{m^2-4n}\right)^p}\left( m+\sqrt{m^2-4 n}\right)^p \\
\frac{\left(m^2-m^2+ 4 n\right)^p}{\left(m+\sqrt{m^2-4 n}\right)^p}+\frac{\left(m^2-m^2+ 4 n\right)^p}{\left(m-\sqrt{m^2-4n}\right)^p} \\
\frac{2^{2p}n^p}{\left(m+\sqrt{m^2-4 n}\right)^p} + \frac{2^{2p}n^p}{\left(m-\sqrt{m^2-4 n}\right)^p} \\
2^{2p}\left(\frac{n^p}{\left(m+\sqrt{m^2-4 n}\right)^p} + \frac{n^p}{\left(m-\sqrt{m^2-4 n}\right)^p} \right) \\

\]