Conference rusure::math

>    1. Find a number of which the square when increased or decreased by 5
>       remains a square.  (All numbers are to be taken to be rational numbers.)

        Letting the middle number be (a/b)^2 where (a,b) = 1,
        this will be solved by finding positive integers a and b
        with (a,b) = 1, and a^2 - 5b^2 and a^2 + 5b^2 both squares.
        Try a = 41, b = 12 to get (31/12)^2, (41/12)^2, (49/12)^2
        which differ by 5.
        
        Dan

>    2. Generalize.  (Find all sequences of three squares in arithmetic
>       progression.)

        Let the squares be the squares of the three rationals
        x-a, x, and x+b, with the difference between the squares
        being t:
        
        	(x-a)^2 = x^2 - t
        	x^2     = x^2
        	(x+b)^2 = x^2 + t
        
        Solving these for x yields x = (a^2 + b^2) / (2(a-b))
        after which one can solve for t to get t = ab(a+b) / (a-b)
        
        All sequences of three squares of rationals in arithmetic
        progression can be gotten by letting a and b vary over
        pairs of unequal rationals, and computing x and the
        squares as above.  The (rational) difference between the
        squares is t.  (Use a=5/6, b=2/3 to get the example I
        posted in .-1 for t=5.)
        
        Dan

>    3.  Finds all sequences of four squares in arithmetic progression.
        
        None of the three thousand sequences of three squares in
        arithmetic progression which I had a program randomly
        generate could be extended on either side to be a
        sequence of four squares in arithmetic progression.
        
        Dan