Let and . The second equation transforms to . The first equation transforms to
Next, noting the expansion for the cube of a sum gives us
Since we know , this simplifies to
Now we have the system of equations
Subtracting the second one from the first one gives us . Since and , and are some permutation of and . If and , then so or . If and , then so and are the roots of the equation , which has a negative descriminant. Therefore, the only solutions are .

Let be real numbers such that



Now, we determine the coefficients through one by one. The sum of the coefficients is , so we have , which means . The sum of the coefficients is also , so we can say that which after plugging in gives . By repeating this same pattern, we can determine that , ,... these seem like alternating signs of the Fibonacchi sequence. After proving this, we compute the first fibonacchi numbers besides the first ) to find , which we know is positive since it is the coefficient of a term with odd degree:

Since the coefficient of is the same as the coefficient of , we are done.