1We have to prove that , m = n and mn = k 2 + k + 3 holds for atleast 1 m , n and k , each being a +ve integer .
Or , k 2 + k + 3 - m 2 = 0 for some k and some m .
We check the discriminant .
1 - 4 ( 3 - m 2 ) ≥ 0 --------------- THIS MUST HOLD FOR SOME +VE m .
Or , 4 m 2 - 11 ≥ 0 --------- for some +ve m .
But that holds for any m ≥ 2 .
So such a possibility exists .
I still think something is wrong with the question , this certainly is not an Olympiad level question, is it ??????
3SORRY BHAIYYA BUT , THE QUESTION IS PERFECTLY CORRECT .
11(x^2+11y^2)^2=16mn=16(k^2+k+3)
Now for checking whether the r.h.s can be a perfect square or not, let k^2+k+3=a^2
Since discriminant is a perfect square, we have 4a^2-t^2=11 for some natural t.
Which gives 1 of the values of a as 3.
Now lets see whther (x^2+11y^2)^2=16a^2 admits any odd valued pair of integers with a=3.
from which we have (x,y)=(1,1) as a soln.
So abt m and n we have 16mn=16.9 thus assuming m\ne n we have (m,n)=(9,1)
x^2+11y^2=36 does have a soln (maybe unique) (x,y)=(5,1).
Done.
