Question

In the light of my 3 previous problems, this will be very easy to do. Let m,n be integers and p be an integer greater or equal to zero, then
\[\left(m-\sqrt{n}\right)^p+\left(m+\sqrt{n}\right)^p
\]
is an integer.

Answer 1

yes

Answer 2

All you need is this problem
http://openstudy.com/users/eliassaab#/updates/54998a46e4b0b8a54fae714e

Answer 3

yes the coefficient are alternative so
the square roots disapears

Answer 4

m-sqrt n is real let it be r
m+sqrt n is real let be be s
done

Answer 5

I think this should work when \(p\) is even :
\[(m-\sqrt{n})^p + (m+\sqrt{n})^p = \sum \limits_{k=0}^{p/2} \binom{2p}{2k} m^{p-k}n^{k} \in \mathbb{N}\]

Answer 6

binomial thm is too irresistable to not think about ;p

Answer 7

that remind me of number theory question
(m-sqrt n )^p+ (m+ sqrt n)^p=0 mod something

Answer 8

Please, note that I think the solution can be given using the induction principle on integer p, namely:
for p=1, we have:
\[(m-\sqrt{n})+(m+\sqrt{n})= 2m\]
so it is true,
for p=2, we have:
\[(m-\sqrt{n})^{2}+(m+\sqrt{n})^{2}=m ^{2}+n-2m \sqrt{n}+m ^{2}+n+2m \sqrt{n}=2(m ^{2}+n)\]
so it is true,
Now I suppose that your statement it is true for an integer p-1, so it is true also for integer p-2.
namely:
\[(m+\sqrt{n})^{p-1}+(m+\sqrt{n})^{p-1}\]
is an integer.
Now, for p, we note that:
\[[(m-\sqrt{n})^{p-1}+(m+\sqrt{n})^{p-1}][(m+\sqrt{n})+(m-\sqrt{n})]=\]
\[(m-\sqrt{n})^{p}+(m+\sqrt{n})^{p}+(m-\sqrt{n})^{p-1}(m+\sqrt{n})+\]
\[+(m-\sqrt{n})(m+\sqrt{n})^{p-1}\]
which is an integer by induction hipothesis.
Now we can observe that:

Answer 9

\[(m-\sqrt{n})^{p-1}(m+\sqrt{n})+(m-\sqrt{n})(m+\sqrt{n})^{p-1}=\]

Answer 10

\[=(m-\sqrt{n})^{p-2}(m-\sqrt{n})(m+\sqrt{n})+(m+\sqrt{n})^{p-2}(m-\sqrt{n})(m+\sqrt{n})=\]

Answer 11

\[=[(m-\sqrt{n})^{p-2}+(m+\sqrt{n})^{p-2}](m-n)\]
which is an integer by induction hypotesis, so our thesis is demonstrated!

Answer 12

ooops.. \[[...](m ^{2}-n)\]

Answer 13

our thesis is true because:
\[(m-\sqrt{n})^{p}+(m+\sqrt{n})^{p}\]
is expressed as difference between two integers!

Answer 14

Since m is an integer, it doesn't matter what power it's raised to.

When -sqrt(n) is raised to an even power it removes the square root and becomes an integer and it doesn't matter.

When -sqrt(n) is raised to an odd power it cancels out with its counterpart in the other term on the right and it doesn't matter.

box proof thingy here.