### Roots and Coefficients

If xyz = 1 and x+y+z =1/x + 1/y + 1/z show that at least one of these numbers must be 1. Now for the complexity! When are the other numbers real and when are they complex?

### Target Six

Show that x = 1 is a solution of the equation x^(3/2) - 8x^(-3/2) = 7 and find all other solutions.

### 8 Methods for Three by One

This problem in geometry has been solved in no less than EIGHT ways by a pair of students. How would you solve it? How many of their solutions can you follow? How are they the same or different? Which do you like best?

# Sextet

##### Stage: 5 Challenge Level:

Oliver from Olchfa School and Simon from Elizabeth College, Guernsey both proved a general property of the sequence, namely that each term is the difference of the two previous terms. From this they found that the sequence is a repeating cycle of six values. You may like to consider more general sequences $a, b, b-a, ...$ with the property that each term is the difference of the two previous terms and investigate whether such sequences are always cyclical.

This is Oliver's solution:

Let $A_1 = x + {1\over x} = 1$ and note that $x$ has no real solutions. Let $A_n = x^n+ {1\over x^n}$.

We have $A_0 = 1 + 1 = 2$.

For $A_2 = x^2+ {1\over x^2}$, as $(x + {1\over x})^2 = x^2 + 2 + {1\over x^2}= 1$ so $A_2 = A_1 - A_0 = 1 - 2 = -1$.

For $A_3 = x^3+ {1\over x^3}$, since $A_2 = x^2 + {1\over x^2} = (x^2 + {1\over x^2})(x + {1\over x}) = x^3 + x + {1\over x} + {1\over x^3} = A_3 + A_1$, therefore $A_3 = A_2 - A_1 = -1 -1 = -2$.

In general $A_{n-1} = x^{n-1} + {1\over x^{n-1}} = (x^{n-1} + {1\over x^{n-1}})(x + {1\over x}) = x^n + x^{n-2} + {1\over x^n} + {1\over x^{n-2}} = A_n + A_{n-2}$, therefore $A_n = A_{n-1} - A_{n-2}$.

From $A_0=2$ and $A_1=1$ we can generate the whole sequence of $A_n$ as follows: 2, 1, -1, -2, -1, 1, 2, 1, -1, ... We can see that the sequence is a repeating pattern of 2, 1, -1, -2, -1, 1 for successive values of $n$ with a period of 6.

Rupert from Wales High School noted that:

$x^3= -1$ and hence that $x^4 = -x$, $x^5 = -x^2$ and $x^6 = -x^3 = 1$. If $x + {1\over x}=1$ then $x^2-x + 1 = 0$ and $x\neq -1$ so $(x+1)(x^2 - x +1)=x^3 +1 =0$. Hence $x$ is the cube root of $-1$ so $x^4 = -x$ and $x^7 = x$. From this it is easy to show that $x^n + {1\over x^n}$ takes the values 1, -1, -2, -1, 1, 2, 1, -1, .... cyclically.