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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?

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    Is there a typo error? You wrote $m^7+n^7+30=1$. Is this what you mean? – 2017-01-24
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    $\displaystyle\text{Sorry, the question has some errors, should be: }\ {m}^{2}={1}-{m},{\quad\text{and}\quad}{n}^{2}={1}-{n}$ – 2017-01-24

4 Answers 4

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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 $m^7 + n^7 = 1$.

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    I haven't found the roots explicitly. – 2017-01-24
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    Not only that, he asked to prove $m^7+n^7\mathbf{+30}=1$. – 2017-01-24
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    That is not correct. The 30 in the equation must be a mistake (since we get $m^7 + n^7 = 1$) – 2017-01-24
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    Sorry, should be $\displaystyle{m}^{2}={1}-{m},{\quad\text{and}\quad}{n}^{2}={1}-{n},{\quad\text{and}\quad}{n}\ne{m};$ – 2017-01-24
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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.

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

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    this is unreadable. To render mathematics text, just enclose the text in dollar signs. – 2017-01-24
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    You can get some hints from the GrindEQ output, but you really have to be careful because there is more bad formatting there than good formatting. Partly this is because it uses a lot of LaTeX commands that simply don't work on this site, partly it is due to the idiosyncrasies of MS Word. – 2017-01-24
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    Your question says $m^2=m-1\,$. Your answer starts with $x^2=1-x\,$. Which one do you mean? – 2017-01-24
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    $\displaystyle\text{Sorry, the question has some errors, should be: }\ {m}^{2}={1}-{m},{\quad\text{and}\quad}{n}^{2}={1}-{n}$ – 2017-01-24
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[ 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 $m^7+n^7=-29\,$.

[ 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:

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