Discussion @ 1335. White Thesis @ Timus Online Judge

manishmmulani (n*n) (n*n + n + 1) (n*n + 1) =====> O(1) soln [6] // Problem 1335. White Thesis 6 Jan 2008 02:34

Rustambek_UWED Re: (n*n) (n*n + n + 1) (n*n + 1) =====> O(1) soln [2] // Problem 1335. White Thesis 7 May 2008 17:44

Thank you very much for the formula. I can prove it. But had not found it.

Desperados[KhAI] >> Starov_Lesha Re: (n*n) (n*n + n + 1) (n*n + 1) =====> O(1) soln [1] // Problem 1335. White Thesis 15 Jun 2008 15:58

Something very dark...
The solution is more easy and more easy to understand too!

Denis Koshman Re: (n*n) (n*n + n + 1) (n*n + 1) =====> O(1) soln // Problem 1335. White Thesis 24 Jul 2008 16:09

n^2+n, n^2+2n, n^2

Megatron Re: (n*n) (n*n + n + 1) (n*n + 1) =====> O(1) soln [2] // Problem 1335. White Thesis 10 Mar 2009 10:49

I can prove it . but i don't know how did you come out with that?

IgorKoval(from Pskov) Re: (n*n) (n*n + n + 1) (n*n + 1) =====> O(1) soln [1] // Problem 1335. White Thesis 3 Dec 2011 03:37

Megatron wrote 10 March 2009 10:49

I can prove it . but i don't know how did you come out with that?

It's simple.
Just c=n^2 ( by intuition ). So, find a and b.
a=n^2+c and b=n^2+d ( because n^2 <= a,b,c <= (n+1)^2 = n^2 + 2*n + 1 ), where c and d is positive and c!=d( because a!=b(by text of this problem ) )

a^2 + b^2 = (n^2+c)^2 + (n^2+d)^2 = 2*n^4 + 2*n^2(c+d) + c^2 + d^2

Look at this 2*n^4 + 2*n^2(c+d) + c^2 + d^2. Each summand must be divisible on c=n^2.
So, just, c = n and d = 2*n
Answer:
a = n^2 + n
b = n^2 + 2*n
c = n^2

P.S.:
You can say just a = n^2 and analogous prove another formula.

Edited by author 03.12.2011 03:41

LyanA Re: (n*n) (n*n + n + 1) (n*n + 1) =====> O(1) soln // Problem 1335. White Thesis 21 Jan 2013 20:11

i was very surprised
it's just 3 strings of code
you're perfect =)

Discussion of Problem 1335. White Thesis