A proof of an equation

[ Proposition ]

 

For such an equation, x^n + a(1) * x^(n-1) + ... + a(n) = 0, where n is a positive integer and a(i) is constant. Once it has rational solutions, therefore the solutions must be integer.

 

[ Solution ]

 

Assume that x = p / q, then we can change the form of equation like this:  sum { a(i) * (p / q) ^ (n - i) } = 0.

We have

 

sum { a(i) * (q ^ i) * (p ^ (n - i)) } = 0, where a(0) = 1, 0 <= i <= n.

 

Then

 

sum { a(i) * (q ^ i) * (p ^ (n - i)) } = - (q ^ 0) * (p ^ n) = - (p ^ n), where 0 < i <= n.

 

As we known

 

gcd(q ^ 1, q ^ 2, ..., q ^ n) | (- (p ^ n))  

 

[ As for equation: sum{ a(i) * x } = y (1 <= i <= n), if it has solutions, then gcd(a(i) | 1 <= i <= n) | y.  ]

 

That's mean

 

q | p.

 

So x = p / q must be an integer.

 

 

 

 

2011-03-04 by hzwu.