Progressions with perfect squares! (Posted on 2007-12-20)

	Submitted by Chesca Ciprian
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First, is easy to prove, in both case (arithmetical or geometrical progression) that because all elements are natural numbers, the ratio is a natural number too. Let note r = ration; a(n) = the n elements for arithemtical progression; b(n) = the n elements for geometrical progression; p = any natural number; a^2 = first element; Prove for arithmetical progression : If we take n = p^2*r+2*p*a+1 we find that a(n) = a^2+r*(p^2*r+2*p*a+1-1) = a^2+p^2*r^2+2*p*a*r = (p*r+a)^2 a perfect square. Prove for geometrical progression : If we take n = 2*p+1 then b(n) = a^2*r^(n-1) = a^2*r^2*p = (a*r^p)^2 a perfect square. Because p can be any natural number there are a infinity elements who are perfect square!

	Subject	Author	Date
	Solution	Praneeth	2007-12-26 01:10:42
	complete solution	Daniel	2007-12-20 12:21:09