Proof without finding the roots of a quadratic equation

Question

A question:

If $\displaystyle{m}^{2}={1}-{m},{\quad\text{and}\quad}{n}^{2}={1}-{n},{\quad\text{and}\quad}{n}\ne{m};$

Proof that $\displaystyle{m}^{7}+{n}^{7}+{30}={1}$

Without finding the roots of equation $\displaystyle{x}^{2}+{x}-{1}={0}$.

Is there such a shortcut solution?

Answer 1

We have $m+n = 1, mn = 1$ since $m,n$ are roots of $x^2 - x + 1 = 0$. Now, $m^2 + n^2 = (m+n)^2 - 2mn = -1$, $m^4 + n^4 = (m^2+n^2)^2 - 2m^2n^2 = -1$, $m^3 + n^3 = (m+n)^3 - 3mn(m+n) = 1 - 3 = -2$ and finally, $$2 = (m^4+n^4)(m^3+n^3) = m^7 + n^7 + m^3n^3(m+n) = m^7 + n^7 + 1$$ and hence $m^7 + n^7 = 1$.

Answer 2

Note: The OP has changed the equation of the question. I am not going switch my answer accordingly because the logic remains the same.

Re-write the equation as $x^2 = x – 1$.

Then, $x^4 = (x – 1)^2 = … = - x$.

Also, $x^6 = (x – 1) ^3 = … = x^3 + 2$

∴ $x^7 = x^4 + 2x = (-x) + 2x = x$

∴ $m^7 + n^7 = m + n = 1$ by sum of roots.

Answer 3

Suppose $x^2=1-x$, Equation 1

Multiply each side by $x$,

$x^3=x-x^2$

$=x-(1-x)=2x-1.$

Then $x^6=(2x-1)^2=4x^2-4x+1=4(1-x)-4x+1=5-8x.$

Then $x^7=x x^6=x(5-8x)=5x-8x^2=5x-8(1-x)=13x-8.$

Thus consider the 2 roots of Equation 1, $m$ and $n$, whereby $m+n=-1$, then

$m^7+n^7=(13m-8)+(13n-8)=13(m+n)-16=13(-1)-16=-29.$

Finally, $m^7+n^7+30=-29+30=1.$

Answer 4

[ EDIT ] Hint for the just-edited question, which now has $x^2+x-1=0\,$.

By Vieta's $mn=-1$, so $m^7+n^7=m^7-\cfrac{1}{m^7}\,$.

Dividing the equation by $x \ne 0$ gives $x - \cfrac{1}{x} = -1\,$, then successively:

$$ \require{cancel} -1 = \left(x - \cfrac{1}{x}\right)^3 = x^3 - \cfrac{1}{x^3}- 3 \left(x - \cfrac{1}{x}\right) = x^3 - \cfrac{1}{x^3} + 3\\ \;\;\implies\;\; x^3 - \cfrac{1}{x^3} = -4 $$

$$ -1 = \left(x - \cfrac{1}{x}\right)^5 = x^5 - \cfrac{1}{x^5}- 5\left(x^3 - \cfrac{1}{x^3}\right)+10\left(x - \cfrac{1}{x}\right) = x^5 - \cfrac{1}{x^5} +20-10\\ \;\;\implies\;\; x^5 - \cfrac{1}{x^5} = -11 $$

$$ \begin{align} -1 = \left(x - \cfrac{1}{x}\right)^7 & = x^7 - \cfrac{1}{x^7}- 7\left(x^5 - \cfrac{1}{x^5}\right)+21\left(x^3 - \cfrac{1}{x^3}\right) -35 \left(x - \cfrac{1}{x}\right) \\ & = x^7 - \cfrac{1}{x^7} + 77 - 84 + 35 \end{align} \\ \;\;\implies\;\; x^7 - \cfrac{1}{x^7} = -29 $$ Therefore $m^7+n^7=-29\,$.

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[ NOTE ] What follows is the answer to the question as originally posted, which had $x^2=x-1\,$.

$m,n$ are the distinct roots of $x^2-x+1=0\,$ so $m+n=1$ by Vieta's formulas. Multiplying the equation by $x+1 \ne 0$ gives $x^3+1=0\,$ therefore $m^3=n^3=-1$. It follows that:

$$m^7+n^7=\left(m^3\right)^2\,m + \left(n^3\right)^2\,n=(-1)^2\,m+(-1)^2\,n=m+n=1$$