Question

Pick a problem: https://projecteuler.net/problems and let's solve it. Add me in Project Euler, here's my friend key:

515695_59af051bf075df0d0763d75d3df7cf69

Answer 1

I'm just curious, what is this?

Answer 2

and I pick this one :) https://projecteuler.net/problem=492

Answer 3

Define the sequence a1, a2, a3, ... as:

a1 = 1
an+1 = 6an2 + 10an + 3 for n ≥ 1.
Examples:
a3 = 2359
a6 = 269221280981320216750489044576319
a6 mod 1 000 000 007 = 203064689
a100 mod 1 000 000 007 = 456482974

Define B(x,y,n) as ∑ (an mod p) for every prime p such that x ≤ p ≤ x+y.

Examples:
B(109, 103, 103) = 23674718882
B(109, 103, 1015) = 20731563854

Find B(109, 107, 1015).

Answer 4

so the biggest possible prime is 100999993

Answer 5

number theory?

Answer 6

smallest is 109

Answer 7

Yeah, it seems straight forward more or less, I wonder what the trick is since there are a lot of primes going on there haha.

Answer 8

I know that any prime that can't be expressed as a sum of squares is a Gaussian Prime... and we also need 3 mod 4 for this to happen

Answer 9

hmmm

Answer 10

I think we can turn this recursive formula into a formula for the nth term.

Fibonacci chick should know how to do this, because you do something very similar with a quadratic and the golden ratio to find the nth fibonacci number.

Answer 11

O_O! @Alchemista HI! Long time no see

Answer 12

lol you overestimate my abilities based on my name :) My number theory class was absolutely atrocious so I'm a novice here

Answer 13

Hello

Answer 14

@Alchemista do you know how to take the Inverse Fourier Transform? I'm just stuck at that last step for the heat equation x.x!

Answer 15

Personally I would use a database to get the primes in between then put those in a database and run the slogorithm

Answer 16

then once the number has been found, we'd need the chinese remainder thm

Answer 17

Chinese Remainder Theorem is fun... but I think there are two versions of it... at least that's what I remember in the book.

Answer 18

I know there is a group theory version and a number theory option

Answer 19

I've done it before... just need to refresh my memory on it XD. I think I need to find my professor's notes again.

Answer 20

I mean if we say there are n primes between the two, we want something like this
\[m_1=xmod p_1\]\[m_2=xmod p_2\].
.
.
\[m_n=x mod p_n\]

Answer 21

then we can say \[\frac{m_1}{x}=1modp_1\]\[\frac{m_2}{x}=1modp_2\].
.
.
\[\frac{m_n}{x}=1modp_n\]

Answer 22

right?

Answer 23

oh wait no...

Answer 24

set all \(m_i=m\)

Answer 25

and make it \(x_i\) that are not necessarily distinct