**#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.

1st question is equivalent to (n+12)^2-123 is a pf. square, (n+12)^2=k^2+123,

ReplyDeleteso 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

mayank