## moneybird 3 years ago Find all integers n such that (28.5)^n + (99.5)^n is an integer.

1. Ishaan94

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

2. moneybird

hint: express the decimals into fractions

3. moneybird

they are odd integers, but not all of them

4. Ishaan94

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

5. moneybird

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

6. Ishaan94

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

7. moneybird

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

8. Ishaan94

Yeah...

9. moneybird

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

10. Ishaan94

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

11. Ishaan94

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

12. moneybird

ok i got to go

13. Ishaan94

Bye, Thanks for the Help :-)

14. asnaseer

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:$n=0,1,3,5,7$

15. moneybird

Good job

16. asnaseer

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

17. moneybird

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

18. asnaseer

nice proof - a lot simpler