Equation : (x + m)^3 = n*x

 
 

Consider,     (x + m)^3 \; = \; n \, x    ……….   (1)

where   m   and   n   are positive integers, and the equation (1) has exactly three distinct integer solutions in   x .
 

Let’s take, for example,   m = 30

(x + 30)^3 \; = \; n \, x

possible solutions:  
n \; = \; 6859 \; ( \,= 19^3 \,) ,
n \; = \; 8575 \; ( \,= 5^2\times \; 7^3 \,) ,    and
n \; = \; 29791 \; ( \,= 31^3 \,)
 

n \; = \; 6859    —–>    (x + 30)^3 \; = \; 6859 \, x

then,

x \; = \; -125 \; = \; -5^3
x \; = \; 8 \; = \; 2^3
x \; = \; 27 \; = \; 3^3
 
n \; = \; 8575    —–>    (x + 30)^3 \; = \; 8575 \, x

x = -135 = (-5)\times \; 3^3
x = 5 = 5\times \; 1^3
x = 40 = 5\times \; 2^3
 
n \; = \; 29791    —–>    (x + 30)^3 \; = \; 31^3 \, x

x \; = \; -216 \; = \; -6^3
x \; = \; 1 \; = \; 1^3
x \; = \; 125 \; = \; 5^3
 

gcd(30, 6859) \; = \; 1
gcd(30, 8575) \; = \; 5
gcd(30, 29791) \; = \; 1
 

If we add the condition that gcd(m, n) \; = \; 1 ,   we end up with two solutions.

 
Can we say that there exist infinitely many pairs   ( \,m, n \,)   satisfying the given conditions?

 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 

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About benvitalis

math grad - Interest: Number theory
This entry was posted in Algebra, Number Puzzles and tagged . Bookmark the permalink.

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