In the case you have given, you can note that since the order of the equation is odd, it will have at least one real root. This is a consequence of the fact that if the polyomial has real coefficients, then then the non-real roots appear as conjugate pairs.

 

Now if you differentiate f(x) you get

 

This means the function is strictly increasing. So, it can have only one real root and obviously that root has to be negative(since f(0) = 1>0 and f is an increasing function)

 

Next, use Rational Roots Theorem (look it up on Wikipedia). In this case the only rational roots possible are 1, -1. These dont satisfy the equation. So the equation has no rational roots.

 

In this way, you have a huge arsenal of methods to attack such problems

or note that   

 

and  

Hint:This is an AGP

 

Let S = 1+3a+5a²+...+(2n-1)a^n-1

 

    aS =     a+3a²+.....+(2n-3)a^n-1 + (2n-1)aⁿ

 

So S(1-a) = 1+2(a+a²+a³+...+a^n-1) - (2n-1)aⁿ

 

Now aⁿ-1=0  1+a+a²+...+a^n-1 = 0 as a1

 

PS: sorry raul, someone had asked for help. didnt notice your post

 

edit: the exponent has been corrected after raul pointed it out

First look at

 

It attains a minimum at x=1 and a maximum at x=-1.

 

 

Drawing an approximate graph at this point will help. You can see that if either or if , then only one real root exists.

 

Hence it is necessary and sufficient that   for the given function to have exactly one real root.

The only solutions are permutations of (5,1,1)

 

Given:

 

Hence

 

Since the exponents are all greater than or equal to zero, the LHS is a product of integers.

 

Since 5 is a prime, one of the factors has to be 5 and the others 1.

 

Suppose

 

Obviously x = 5 and y = z = 1. In this way, we can see that the solutions are permutations of (5,1,1)

Ok I filched from another forum. You can check out the other solutions here: http://www.mathlinks.ro/viewtopic.php?p=1089628#1089628

anyone else?