Consider three positive integers x< y< z in arithmetic sequence.
Determine analytically all possible solutions of each of the following equations:
(I) x^{3} + z^{3} = y^{3} + 10yz
(II) x^{3} + y^{3} = z^{3}  2, with y< 116.
Let
a represent the sequence separation where
x=ya and
z=y+a. Equation I can be rewritten:
(ya)³  y³ + (y+a)³ = 10y(y+a)
(y³3y²a+3ya²a³)  y³ + (y³+3y²a+3ya²+a³) = y(10y+10a)
y(y² + 6a²) = y(10y+10a)
(y² + 6a²) = (10y+10a)
y(y  10) = a(6a+10)
Setting the terms inside the parentheses equal to zero yields
y=10, a=10/6. This means if y<10, then a>10/6 (or a>=2) and if y>10, then a<10/6 (or a=1).
First, let's consider
y>10. In that case,
a=1. So solving for
y:
(y³3y²+3y1)  y³ + (y³+3y²+3y+1) = y(10y+10)
y²  10y + 4 = 0
y = 5 ± √21
Since
y does not equal an integer,
y cannot be greater than 10.
Solving for
y<10, minimize the left side of the equation by taking the derivative:
d[y(y  10)]/dy = 2y10 = 0  y=5, y(y  10)=25
Setting the left side equal to this minimum value yields:
a(6a+10) = 25
6a²10a25 = 0
a = (5/6) [1±√7] or a = 3.038
We now know the boundaries of
a whereas
10/6 < a < 3.038. Therefore the solution is now limited to the following:
y = {3,4,6,7,8,9}, a = {2,3}
Solving the right side for both values of a yields:
a=2, a(6a+10)=(4)
a=3, a(6a+10)=(24)
We can now express the right side as follows:
a=2, y(y  10)=(4)  y²10y+4=0
a=2, y(y  10)=(24)  y²10y+24=0
The first equation (a=2) has no integer roots. However, the second equation (a=3) factors (y4)(y6)=0. Therefore,
y=4 or
y=6.
The integer solutions are:
x=1, y=4, z=7 or x=3, y=6, z=9
Edited on February 28, 2007, 1:35 pm

Posted by hoodat
on 20070228 13:33:54 