Getting Perfect Squares With 4 And 9? (Posted on 2007-09-03)

	Submitted by K Sengupta
Rating: 4.0000 (6 votes)
Solution:	(Hide)
If possible, let us suppose: 4*10^p + 9 = q^2 or, 4*10^p = (q+3)(q-3) Let us substitute (P, Q) = (q+3, q-3) such that r = gcd( P, Q) Now, P- Q = 6, and so that r must divide 6. However, PQ = 4*10^p is not divisible by 3 for integer p. Since P and Q possess the same parity, this is feasible iff both P and Q are even. Consequently, r=2 Accordingly, PQ = 2^(p+2)*5^p with gcd(P, Q) = 2 and P> Q Therefore, we must have the following cases: Case (i): P = 2^(p+1)*5^p; Q = 2 This gives, P- Q = 6 = 2(10^p -1) Or, 3 = 10^p – 1, which is a contradiction. Case (ii): P = 2*5^p; Q = 2^(p+1) This gives, 6 = P-Q Or, 6 = 2(5^p – 2^p) Or, 3 = 5^p – 2^p This is feasible only when p=1, which is a contradiction since p ≥ 2 Consequently, 4*10p + 9 can never be a perfect square for integers p ≥ 2

	Subject	Author	Date
	re: Solution	C W Gardner	2007-10-03 23:34:35
	Solution	Praneeth	2007-09-04 02:20:34
	If p is even	Brian Smith	2007-09-03 14:24:15