If p is even, p=2*k(k≥1). Then 4*10^p = (2*10^k)²
The smallest square greater than (2*10^k)² is (2*10^k+1)²
Difference=4*10^k+1 ≥ 41 (Clearly greater than 9)
So, no solution for this case.

If p is odd, p=2*k+1(k≥1).
Let c²=40*10^2k + 9
=> (c-3)(c+3)=(2^(2k+3))*(5^(2k+1))
Let RHS of above eq be x. So, we have to find 2 divisors
of x and their product is x such that their difference is 6.
So, They should be of the form
2^(2k+2)*5^a and 2*5^b such that a+b=2*k+1. Both are
multiples of 10, so a=0.
Finally they should be of form
2^(2k+2) and 2*5^(2k+1)
=> difference of 5^(2k+1) and 2^(2k+1) should be 3
which is impossible for k>0.
So, no solutions for this case also.

So, there are no integral values for p(≥2) for which 
4*10^p+9 is a perfect square.  

Edited on September 10, 2007, 8:09 am

Posted by Praneeth on 2007-09-04 02:20:34