/sci/ can't solve this

>>8832873
d^5-d^3=d^3(d^2-1)=d^2(d-1)d(d+1)

note that (d-1)d(d+1) is divisible by 3

two cases:

1) d even: in this case d^2 is divisible by 4, so d^2(d-1)d(d+1) is divisible by 4*3

2) d odd: in this case d-1 and d+1 are both even, so (d-1)d(d+1) is divisible by 4. since 4 and 3 are relatively prime, it follows that (d-1)d(d+1) is divisible by 4*3.

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4KB, 655x117px

>>8832895
Pretty good. Now try this one.

>>8832914
do your own homework brainlet

>>8832914
lol. using the previous result, replace d^3 with d^5 to get [math] 12\sum_{d |n} d^5[/math]

>>8832922
>Can't do math problem
>Cover it up by saying it's someone's homework

These are from Chapter 9 of Elliptic Curves: Number Theory and Cryptography. If you can find a course that uses that book that is already on Chapter 9 by this time of the school year I would be impressed.

>>8832928
>Elliptic Curves: Number Theory and Cryptograph

My school has a course w/ that next semester. Taught by the author.

>>8832928
it doesn't matter where the questions are from, you could assign such elementary congruence questions to people who just learned modular arithmetic a week ago and they'd solve it

the second question is trivial given the first anyway

>>8832928
Are you Korean?

>>8832950
you're right. i would much rather have another IQ thread than one about actual math problems.

>>8832873
What? (mod 12) results in a value 0 <= val < 12, so any d^3 > 12 will contradict this. It only works for d = 0,1,2

>>8833037
wrong

http://www.wolframalpha.com/input/?i=d%5E3+%3D+d%5E5+mod+12

>>8833037
That three line symbol is congruence, not equality.
As in shorthand for d^3 (mod 12) = d^5 (mod 12)

>>8833046
Oh okay it could've been more clear if you just wrote d^3 mod 12 = d^5 mod 12, but I guess you math fags have to make up symbols to make yourself feel special

>>8833047
math fags like to keep things short, and not rewrite mod 12 unnecessarily

>>8833047
It's standard notation

>>8833037
a is congruent to b (mod n) if and only if n divides the difference (a-b)

in symbols: a ≡ b (mod n) iff n|(a-b)

Since I'm guessing you are more familiar with programming, this means that a%n == b%n. (Although I believe that many programming languages diverge from the mathematical definition when it comes to negative numbers.)

>>8832895

Wow, your google-fu is impressive.

>>8832973
This. Stop bitching about actual problems being posted unless you prefer when this shithole is flooded with /pol/ in disguise threads.

I've really been wanting to get into number theory, I hear it's fun. I'm not getting my degree in maths though. So what's a good book for a beginner? What are some prerequisites?

Fun!

d5-d3 = d2(d2-1) = d*(d-1) * d*(d+1)

either d or (d+1) is even, so is d*(d+1)*d*(d-1).
In the same way, 3 is a devider of either d, d+1 or d-1.

In the end, d5 - d3 is a multiple of both 2 and 3, so it is a multiple of 12.

d5 - d3 = k*12
=> d5 == d3 [12]

>>8833333
ooohhh

>>8833140
It's not that hard to figure out by yourself man.
Took me 5 minutes.

>>8833333
>>8833333
>>8833333
>>8833333

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>>8833280
https://en.wikipedia.org/wiki/False_dilemma