Question

Find all integers n such that (28.5)^n + (99.5)^n is an integer.

Answer 1

I think it is true for every odd integer and zero, of course! but I am finding it hard to prove it :/

Answer 2

hint: express the decimals into fractions

Answer 3

they are odd integers, but not all of them

Answer 4

One thing I have noticed that if we end up with '.50' as fractional part, We get an Integer as ".5+.5=1"

Answer 5

\[(\frac{57}{2}) ^n + (\frac{199}{2}) ^n \]

Answer 6

Hmm I did the fractions way too

\[\left(\frac{57}{2}\right)^n + \left(\frac{199}{2}\right)^n \implies \frac{(57)^n + (199)^n}{2^n}\]

Now one can easily see it's valid for n = 1 and 0

Answer 7

You know that
57^n + 199^n must be an integer multiple of 2^n

Answer 8

Yeah...

Answer 9

So now you need to prove that whether n is odd or even

Answer 10

Hmm Okay

\[(57)^n + (199)^n = K*2^n\]

Now \(odd^n \) should be always odd.The LHS is always even as odd + odd = even. \(2^n\) is always even too Now for RHS to be even K should be even as well

Answer 11

ugh I need to think now, I messed up somewhere :/

Answer 12

ok i got to go

Answer 13

Bye, Thanks for the Help :-)

Answer 14

If we rearrange the expression as:\[(\frac{57}{2})^n+(\frac{199}{2})^n\]then as @Ishaan94 pointed out, we can go on to get this equation:\[57^n+199^n=\text{constant} * 2^n\]so we need to prove that \(57^n+199^n\) is divisible by \(2^n\).

I then noticed that \(199=256-57=2^8-57\) and took advantage of this to get:\[57^n+(2^8-57)^n\]and using the binomial expansion we get:\[
\begin{align}
(2^8-57)^n&=(2^8)^n-(2^8)^{n-1}.57+…+(-1)^{n-1}.(2^8).57^{n-1}+(-1)^n.57^n\\
&=2^8(2^n-2^{n-1}.57+…+(-1)^{n-1}.57^{n-1})+(-1)^n.57^n\\
\therefore 57^n+(2^8-57)^n&=57^n+2^8(2^n-2^{n-1}.57+…+(-1)^{n-1}.57^{n-1})+(-1)^n.57^n\\
&=57^n+(-1)^n.57^n+2^8(2^n-2^{n-1}.57+…+(-1)^{n-1}.57^{n-1})\\
&=57^n(1+(-1)^n)+2^8(2^n-2^{n-1}.57+…+(-1)^{n-1}.57^{n-1})\\
\end{align}\]the first expression \(57^n(1+(-1)^n)\) is always zero for odd values of 'n' and is not divisible by \(2^n\) for even values of 'n'.
the second expression \(2^8(...)\) is always divisible by \(2^n\) for \(n=0…8\).

so, combining the two conditions, we get the solution as:\[n=0,1,3,5,7\]

Answer 15

Good job

Answer 16

@moneybird - do you have the actual proof for this and does my proof match up to it?

Answer 17

you know that n must be an odd number
because let n be 2m
\[(57^{m})^{2} + (199^{m})^{2} = constant * 2^n\]

odd integer = 2k + 1
square of an odd
(2k + 1)^2 = 4 (k^2 + k) + 1

so sum of two squares of odd must be 2 more than the multiple of 4.

2^n is divisble by for n is greater or equal to 2

So n must be an odd number

\[57^n + 199^n = (57 + 199) (57^{n-1} - 199 *57^{n-2} ... + 199^{n-1})\]

57 + 199 = 256

57^n + 199^n is divisble by 256

\[\frac{256}{2^n} = q\] where q is an integer

n can only be 0, 1, 3, 5 and 7

Answer 18

nice proof - a lot simpler