Perfect square

This is one of the important topics which often comes in various competitive exams.

#Problem 5(A)

Find the total number of values of n for which n^2 + 24n + 21 is a perfect square.Where n is natural number.

Solution scheme and approach
Let n^2+24n+21 is a perfect square.Hence we can write
n^2+24n+21=k^2
=>n^2+2*n*12+12^2-12^2+21=k^2
=>(n+12)^12-k^2=123
Case I
(n+12+k)(n+12-k)=3*41=(-3)*(-41)
So, (n+12+k)+(n+12-k)=3+41
=>2(n+12)=44 =>n=10
or,
2(n+12)=-44=>n=-34
Case II
(n+12+k)(n+12-k)=1*123=-1*-123
So, 2(n+12)=124=>n=49
or,
2(n+12)=-124=>n=-73
Therefore total 4 values of n ::{-73,-34,10,49} are possible.
Food for thought

Find the total number of values of n for which n^2+24n+21 is square of a prime number.Where n is natural number.

What would be the answer then.Understand the difference

#Problem 5(B) 
Find all positive integers n such that n^2 + n + 2009 is a square.[Pomona-Wisconsin Math Talent Search 2009]
Solution scheme and approach
Let, n^2+n+2009=k^2
=>(n+1/2)^2-k^2=1/4-2009[Follow the same method as stated in 5(A)]
=>(n+1/2+k)(n+1/2-k)=-8035/4
=>[2*(n+1/2+k)][2*(n+1/2-k)]=-8035
 (PS:As n is an integer we need to maintain the symmetry. So don't consider [4(n+1/2+k)]*(n+1/2-k) or (n+1/2+k)*[4(n+1/2-k)]. Each of the cases LHS becomes fraction hence contradicts)
 =>(2n+1+2k)(2n+1-2k)=-8035
Now as 2n+1+2k>2n+1-2k
therefore, (2n+1+2k)(2n+1-2k)=1607*-5=8035*-1
=>2(2n+1)=1607-5=1602
=>n=400
or,
2(2n+1)=8034=>n=2008
Two values of n are possible.