11Varun i tried it by applying no theory but on trying further i got stuck on modular equation which could not be simpified further.
By the way have you tried to prove their unique existence by contradiction.(although i could not get it)
prove that every integer can be expressed in this form and also for a partical integer there exists unique values of a and b
n=1/2(a+b-1)(a+b-2) +a
my sol i solved it involes something related to consecutive numbers and remainders
can anyone hv an application of number theory 2 solve it or if he is a real genius inequality.....????
-
UP 0 DOWN 0 0 3
3 Answers
111i poted the whole sol n the server timed out ....
always substitue values n generate ideas
simple let
x(x+1)<2n<(x+1)(x+2)
then
2n-2a=x(x+1) (substit. the value of a+B-_
now
2n-x(x+1)=2a
where 2a=2,4,6...2x ( simple as both x(x+1) is even)
now then a=1,2,3,...x
now if x=a+b-1 the minimum value that x can take is 1 then eq. is satisfied( we dont care what b is)
max a=x then b can be 1 thereofre proven and integer
now for
2n=x(x+1)
therefore let a+b-1=x
therefore
x(x+1)-x(x-1)=2a
thereofre a=x and imiarly a nad b can pe proven .....
V imp we hv proven that any integer can pe proven like this not any integer 2n.......
y coz we know the values of a and b. so we can express n as form of that ..... 2n=(a+b-1)(a+b-2)+2a
while n=1/2(a+b-1)(a+b-2) +a
n obv one of a+b-1 or.a+b-2 will be divisible by 2 an we know a
can u go on and prove that it is the only unique sol.. same principle...
P.S. THIS IS MY OWN PROCESS ITS RIGHT I AM SURE MAYBE ITS THE OFFICIAL AND ONLY SOLUTION.I DONT KNOW I HAVE NOT SEEN IT
21isn't q too easy for INMO ?
Let a+b-2 = n ; n\geq 0 . The expression becomes \frac{n(n+1)}{2}+ a ;
Let us define a function f(p) = \frac{p(p+1)}{2} ,whose domain is in non-negative integers. We have f(p+1)- f(p) = p+1 and also 1\leq a\leq n+1. This prove existence and uniqueness at the same time.