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 Diophantine And Almost Fermat (Posted on 2007-02-21)
Consider three positive integers x< y< z in arithmetic sequence.

Determine analytically all possible solutions of each of the following equations:

(I) x3 + z3 = y3 + 10yz

(II) x3 + y3 = z3 - 2, with y< 116.

 See The Solution Submitted by K Sengupta Rating: 4.0000 (1 votes)

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 Solution - Equation I Comment 4 of 4 |
Let a represent the sequence separation where x=y-a and z=y+a.  Equation I can be rewritten:

(y-a)³ - y³ + (y+a)³ = 10y(y+a)

(y³-3y²a+3ya²-a³) - 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:

(y³-3y²+3y-1) - 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 = 2y-10 = 0    |    y=5,  y(y - 10)=-25

Setting the left side equal to this minimum value yields:

a(-6a+10) = -25

6a²-10a-25 = 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)   |   y²-10y+4=0
a=2, y(y - 10)=(-24)  |  y²-10y+24=0

The first equation (a=2) has no integer roots.  However, the second equation (a=3) factors (y-4)(y-6)=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 2007-02-28 13:33:54

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