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Further Arithmetic Integers
K Sengupta asked for integer solutions
to x^4+y^4=z^464 with z>y>x>0 and in arithmetic sequence
My question is concerning the more general problem
x^4+y^4=z^4n with z>y>x>0 and in arithmetic sequence
and for what values of n are there only two solutions
As already established, using x = y  r and z = y + r, the equation x^4 + y^4 = z^4  n becomes
y^4  8r(y^3)  8(r^3) y + n = 0.
Some necessary conditions have already been posted (y divides n, and 8r divides n + y^4), but not sufficient to ensure a solution. For example, n = 40, r = 1, y = 2, satisfy these conditions but not the equation.
Part 1  some thoughts
Dividing by r^4 and writing Y = y/r gives: Y^4  8Y^3  8Y + n/r^4 = 0 (1)
For fixed n and r, this quartic in Y (has one stationary point  a minimum value of n/r^4  480.219.. when Y = 6.0545...(both irrationals). When n = 0 it has roots at Y = 0 and 8.121..
It follows that, for n > 0, there will be two real solutions for Y provided that n/r^4 < 480.219.. (otherwise none) and that these solutions will be in the range 0 < Y < 8.121..
Now, we are looking for rational solutions for Y, so that y and r can be integers, and we can now say:
For a given r value, there will be, at most, two integer values for y and they will satisfy the inequalities y < 8r and n < 480.219 r^4.
This is still very vague, but these limits allow computer searches to be better organized and more conclusive, which, in turn, gave me an insight into the following..
Part 2  some thoughts
Equation (1) can be rearranged to give
(Y  2)(Y  8)(Y^2 + 2Y + 4) + (n  64r^4)/r^4 = 0.
It's clear that when n = 64r^4 there are two rational roots for Y, namely Y = 2 and Y = 8. This means that, for a given r value, our problem will have two solutions, namely y = 2r and y = 8r when n = 64r^2.
E.g. r = 1 gives n = 64, y = 2 or 8 (1, 2, 3 and 7, 8, 9).
r = 2 gives n = 1024, y = 4 or 16 (2, 4, 6 and 14, 16, 18) etc.
So, n = 64r^2 , with r = 1, 2, 3….always gives two solutions However, I can't be sure that it gives all possible values of n with this property…
Edited on July 26, 2009, 7:56 pm

Posted by Harry
on 20090726 19:54:28 