Number Theory 11/26

[Steven G]

Prove that 1^5+2^5+...+100^5=0 (mod 4)

 

[topsquark]

Well, by the Binomial Theorem:
[imath](4n + 0)^5 = 4( \text{some number} ) + 0^5 \equiv 0 \text{ (mod 4)}[/imath]
[imath](4n + 1)^5 = 4( \text{some number} ) + 1^5 \equiv 1 \text{ (mod 4)}[/imath]
[imath](4n + 2)^5 = 4( \text{some number} ) + 2^5 \equiv 0 \text{ (mod 4)}[/imath]
[imath](4n + 3)^5 = 4( \text{some number} ) + 3^5 \equiv -1 \text{ (mod 4)}[/imath]

When we sum from [imath]1^5[/imath] to [imath]100^5[/imath] we are adding the same number of 1 (mod 4) terms and -1 (mod 4) terms, so they cancel.

Thus all the terms that could be not equal to 0 (mod 4) cancel. Thus the sum is equal to 0 (mod 4).

-Dan

 

[Steven G]

Well, by the Binomial Theorem:
[imath](4n + 0)^5 = 4( \text{some number} ) + 0^5 \equiv 0 \text{ (mod 4)}[/imath]
[imath](4n + 1)^5 = 4( \text{some number} ) + 1^5 \equiv 1 \text{ (mod 4)}[/imath]
[imath](4n + 2)^5 = 4( \text{some number} ) + 2^5 \equiv 0 \text{ (mod 4)}[/imath]
[imath](4n + 3)^5 = 4( \text{some number} ) + 3^5 \equiv -1 \text{ (mod 4)}[/imath]

When we sum from [imath]1^5[/imath] to [imath]100^5[/imath] we are adding the same number of 1 (mod 4) terms and -1 (mod 4) terms, so they cancel.

Thus all the terms that could be not equal to 0 (mod 4) cancel. Thus the sum is equal to 0 (mod 4).

-Dan

I used the fact that 2²=32 = 0 (mod 4). As a result of this, (2n)⁵=2⁵n⁵=0 (mod 4) (where n is a positive integer).
Then the odd bases give either 1 or -1 as you noted.

 

[lookagain]

I used the fact that 2²=32

2⁵ = 32