MHB Find n in A=(108.5)^n+(147.5)^n

[Albert1]

$for \,\, n\in N$
$A=(108.5)^n+(147.5)^n \,\, also \in N$
find $all \,\, n$

 

[Albert1]

$for \,\, n\in N$
$A=(108.5)^n+(147.5)^n \,\, also \in N$
find $all \,\, n$

hint:

Spoiler

$108.5=\dfrac {217}{2}, 147.5=\dfrac{295}{2}$
and $217+295=512=2^9$

 

[kaliprasad]

$for \,\, n\in N$
$A=(108.5)^n+(147.5)^n \,\, also \in N$
find $all \,\, n$

Spoiler

We have $A = \frac{217^n + 295^n}{2^n}$
there are 2 cases
n is odd
$217^n+ 295^n = (217 + 295)(217^{n-1} - 216^{n-2} * 295^2 +\cdots 295^{n-1}) = 512(x)$ where x is sum of n numbers each is odd

so $2^n$ must divide $512=2^9$ so n = 1 or 3 or 5 or 7 or 9
for n is even
$217^n + 295^n=295^n -217^n + 2* 217^n = (295 - 217)(217^{n-1} - 216^{n-2} * 295^2 +\cdots 295^{n-1}) + 2* 217^n$

$295^n -217^n$ is divisible by 4 as in the product 1st term is divisible by 2 and 2nd one also 2

but $2* 217^n$ is not
so the sum is not divisible by 4
so there is no even n ( as $2^2= 4$)
so only choices of n are 1,3,5,7,9