A Pythagorean piglet (Posted on 2012-09-22)

	Submitted by broll
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Making the substitution: x^2- b^2=b^2-1 becomes x^2- (yz^2)^2=(yz^2)^2-1, giving x^2=2y^2z^4-1. Writing x^2=2(yz^2)^2-1 reveals the Pellian (technically, Non-Pellian) nature of the task. We now need to select a suitable square z, the smallest being 25=5^2. Although A133204 in Sloane refers to 'Primes p such that the non-Pellian equation x^2-2py^2=-1 is solvable', the same is also true, at least, of their squares. Hence x^2-(25y)^2=(25y)^2-1 has a solution, and indeed recurrent solutions, since equivalently x^2 - 2*(5^4)y^2 =-1, see e.g. here: http://www.alpertron.com.ar/QUAD.HTM. It is thereby readily ascertained that 2*7801^2*(5^4)-1=275807^2, with the power of y a square and of z a 4th power, as was required by the problem. 2373672716981597^2*2*(5^4)-1=83922003724759193^2 is another solution, along the same lines.

	Subject	Author	Date
	re(3): Oinking solution.	broll	2012-12-23 09:12:26
	re(2): Oinking solution.	Jer	2012-09-26 12:41:34
	re: Oinking solution.	broll	2012-09-26 00:42:04
	Oinking solution.	Jer	2012-09-25 16:17:25