Saturday, January 2, 2010

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
Case I
So, (n+12+k)+(n+12-k)=3+41
=>2(n+12)=44 =>n=10
Case II
So, 2(n+12)=124=>n=49
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)]
 (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)
Now as 2n+1+2k>2n+1-2k
therefore, (2n+1+2k)(2n+1-2k)=1607*-5=8035*-1
Two values of n are possible.

1 comment:

  1. 1st question is equivalent to (n+12)^2-123 is a pf. square, (n+12)^2=k^2+123,
    so if 123=2p-1, then p=62,
    so if n+12 =62, then k=61, so for n=50 and n=-74, the expression is a pf. square