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Given two relatively prime integers a and b, is there an easy characterization for when a^2+b^2 is square free?

Edit: The above question proved to be too general. The problem I had in my mind is as follows: given the two sequences $a_n$ and $b_n$ defined by $a_0=b_0=1$, $a_{n+1}=a_nb_n,\ b_{n+1}=a_n^2+b_n^2$, are the $b_n$'s square free and coprime?

edited Mar 2 2010 at 23:39

asked Mar 1 2010 at 20:39

Dee
95●6

	But I think it's easy to count the number of such pairs (a,b), asymptotically. That could give an idea of "how likely" it is for a^2 + b^2 to be squarefree. I don't know if this is of any interest for you. – maks Mar 2 2010 at 8:43
	Unfortunately, no! The problem at hand is the recurrence $u_0=1,\ u_{n+1}=u_n+1/u_n$ which corresponds to the two sequences $a_0=b_0=1,\ a_{n+1}=a_nb_n,\ b_{n+1}=a_n^2+b_n^2$. My conjecture is that the $b_n$'s are square free and coprime, which I am unable to (dis)prove until now! – Dee Mar 2 2010 at 10:55
	The $b_n$ are sequence A073833 in the Online Encyclopedia of Integer Sequences, research.att.com/~njas/sequences/A073833 but none of the information there bears directly on the problems of squarefreeness or coprimality. – Gerry Myerson Mar 3 2010 at 2:21

Franz Lemmermeyer · Accepted Answer · 2010-03-01 20:48:13Z UTC

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No. To the best of my knowledge, it is not even known whether Fermat numbers are squarefree.

answered Mar 1 2010 at 20:48

Franz Lemmermeyer
17.2k●1●38●101

Square free sum of two squares

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