A Doubt From Algebra

let m = (2k+1)

n = (2k+3) ,where k is any positive integer

so mn+1 = (2k+1)(2k+3) + 1 = 4k² + 8k + 4 = 4 (k+1)²

m+n = 4(k+1)

mn+1m+n= (k+1)

so All positive integers can be represented by mn+1m+n for positive inegers m and n

Do it this way

case 1> when m is even and n is odd

so, let m = 2k and n = (2k + 1)

Now, mn + 1 = 2k(2k + 1) + 1 = 4k² + 2k + 1

and, m + n = 2k + 2k + 1 = 4k + 1

Thus, mn + 1m + n = 4k² + 2k + 14k + 1, which is always odd

case II> When m and n are both even

then, mn + 1 will be odd (clearly) and m + n will be even

but, odd/even can never be an integer.

But in the given set, each element is an integer.

Hence, only the odd integers between 0 and 2005 i.e., 1,2,3,........,2004
i.e., 1002 numbers. Ans.

Am i right? Do verify me.

oh! sorry....

I forgot the third case >> When m and n are both odd....!!

@Kunl,

Subho did it all, but only posted the solution which fetched the answer!

ok it was later i realized tht n after tht it said post too old to edit :P