Problem 7: show that 4^n + 4^n+1 + 4^n+2 + 4^n+3 + 4^n+4 is divisible by 11 for any positive integer

If n is even, n

4

+4

n

is divisible by 4

∴ It is composite number

If n is odd, suppose n=2p+1, where p is a positive integer

Then n

4

+4

n

=n

4

+4.4

2p

=n

4

+4(2

p

)

4

which is of the form n

4

+4b

4

, where b is a positive integer (=2

p

)

n

4

+4b

4

=(n

4

+4b

2

+4b

4

)−4b

2

=(n

2

−2b

2

)

2

−(2b)

2

=(n

2

+2b+2b

2

)(n

2

−2b+2b

2

)

We find that n

4

+4b

4

is a composite number consequently n

4

+4

n

is composite when n is odd.

Hence n

4

+4

n

is composite for all integer values of n> 1.

check the picture below.

well, the angle at the vertex d is clearly not a right-angle, let's check for the one at the vertex f which does look like,   however if it's indeed, that means ef is perpendicular to df, in which case their slopes will be negative reciprocal of each other.

so the negative reciprocal of the slope of ef is 1, however the slope of df is not 1, is 1/2, so nope, they're not perpendicular lines.