I will give two math problems that are of interest to me.

These are not new problems, however you will have more fun if you work on them yourself and leave comments on what you find. So if you want to work on it without hints, don't read the comments.

I will post about the answers (not sure I will post THE answers) on Thursday.

1) Let x(1)>0. Let x(n+1) = ( 1 + (1/x(n)) )^n.

For how many values of x(1) does this sequence go to infinity?

2) Find all (x,y) \in N \times N such that x^2+3y and y^2+3x are both squares.

#1 is missing a parenthesis somewhere.

ReplyDeleteFixed, thanks

ReplyDeleteMy idea about #2 ... probably not enough "elementary":

ReplyDeletehttps://www.nearly42.org/vdisk/misc/numby.png

GREAT, THANKS- your solution works, so READERS- if you want to

Deletedo it yourself do not read his the solution pointed to.

For #2: Assume wlog that x>=y. Then x^2+3y < x^2+4x+4 = (x+2)^2,

ReplyDeleteand also x^2+3y > x^2. Since the only perfect square strictly

between x^2 and (x+2)^2 is (x+1)^2, we get x^2+3y=(x+1)^2.

This yields 3y=2x+1, so that y=2a+1 is odd, and so that x=3a+1.

Furthermore y^2+3x = 4a^2+13a+4 =:Q.

Now Q >= (2a+2)^2 = 4a^2+8a+4 and Q < (2a+4)^2 = 4a^2+16a+16.

This implies that either Q=(2a+2)^2 with a=0 and x=y=1,

or that Q=(2a+3)^2 with a=5, x=16 and y=11. Done.