1729 is an awesome number!

First we solve the equation $x + \frac{1}{x} = 1 \Rightarrow x^2 - x + 1 = 0$ The roots of this equation are $x = \frac{1 \pm i\sqrt{3}}{2}$ This root can be written as $x = \cos{\frac{\pi}{3}} \pm i\sin{\frac{\pi}{3}}$ Now let $x = \cos{\frac{\pi}{3}} + i\sin{\frac{\pi}{3}}$ and $\frac{1}{x} = \cos{\frac{\pi}{3}} - i\sin{\frac{\pi}{3}}$

We need to find $x^{1729} + (\frac{1}{x})^{1729}$ . Let it be equal to $S$

Using de moivre's theorem( de moivres theorem ) we get:- $S = (\cos{\frac{\pi}{3}} + i\sin{\frac{\pi}{3}})^{1729} + ( \cos{\frac{\pi}{3}} -i\sin{\frac{\pi}{3}})^{1729}$

$S = 2 \cos{\frac{1729\pi}{3}}$

$S = 2 \cos{(576\pi + \frac{\pi}{3})}$

$S = 2 \cos{\frac{\pi}{3}}$

$S = 2 \times \frac{1}{2} = \boxed{1}$

First, we strike to solve the given equation. Rearranging it, gives $x+\frac{1}{x} = 1 \Rightarrow x^2-x+1=0$ The roots of this equation are $x = \frac{1 \pm \sqrt{3}i}{2}$ The above value should immediately ring a bell if we remember that the $3^{rd}$ root(s) of unity are $\frac{-1 \pm \sqrt{3}i}{2}$ . Let us denote this by $\omega$ that is $\omega = \frac{-1 + \sqrt{3}i}{2} \qquad \text{and}$ $\omega^2 = \frac{-1 - \sqrt{3}i}{2}$ From this, we can immediately make out that $x = -\omega, -\omega^2$ .

But why did I represent $x$ in terms of $3^{rd}$ root of unity?

The answer is simple. Since we are dealing with a large power ( $1729$ ), expressing $x$ in terms of $3^{rd}$ root of unity can help us simplify because of the well-known properties : $1+\omega+\omega^2 = 0$ $\omega^3 = 1$

Also observe that $1728 \equiv 0 \pmod{3}$ .

And thus, we finally left with computing

$\begin{aligned} x^{1729} + x^{-1729} & = & (-\omega)^{1729}+(-\omega)^{-1729} \\ & = & (-\omega)^{1729}+(-\omega^2)^{1729} \\ & = & (-\omega)(-\omega)^{1728}+(-\omega^2)(-\omega^2)^{1728} \\ & = & (-\omega)+(-\omega^2) \\ & = & \boxed{1} \end{aligned}$

Firstly, we see that $1729=1^3+12^3=9^3+10^3$ Let's use $9^3+10^3=729+1000$ Now let's rearrange the given equation: $x+\frac{1}{x}=1 \Rightarrow x^2=x-1$ Then, note that $x^4=(x-1)^2=x^2-2x+1=(x-1)-2x+1=-x$ $x^8=(x^4)^2=(-x)^2=x^2=x-1$ $x^9=(x^8)(x)=(x-1)(x)=x^2-x=(x-1)-x=-1\\ \Rightarrow x^{81} = (-1)^9=-1\\ \Rightarrow x^{729}=(x^{81})^9=(-1)^9=-1$ $x^{10}=(x^9)(x)=(-1)(x)=-x\\ \Rightarrow x^{100}=(-x)^{10}=x^{10}=-x\\ \Rightarrow x^{1000}=(x^{100})^{10}=(-x)^{10}=x^{10}=-x$ Finally, we have $\begin{aligned} x^{1729}+x^{-1729} & = (x^{729+1000})+\frac{1}{x^{729+1000}}\\ &= (x^{729})(x^{1000})+\frac{1}{(x^{729})(x^{1000})}\\ &=(-1)(-x)+\frac{1}{(-1)(-x)}\\ &=x+\frac{1}{x} = \boxed{1} \end{aligned}$ As an aside: A short story behind 1729 .

$\begin{aligned} x + \frac 1 x & = & 1 \\ x^2 + 1 & = & x \\ x^2 - x + 1 & = & 0 \\ (x+1)(x^2-x+1) & = & = 0 \\ x^3+ 1 & = & 0 \\ x^3 & = & -1 \\ \end{aligned}$

Knowing that $1729$ is a taxicab number, therefore $179 = 1^3 + 12^3 = 9^3 + 10^3$

$\Rightarrow x^{1729} = x^1 \cdot x^{12^3} = x \left ( (x^3)^4 \right)^{144}$

With $x^3 = -1 \Rightarrow x^{1729} = x \cdot \left ( (-1)^{144} \right ) = x$

$\Rightarrow x^{1729} + \frac 1 {x^{1729}} = x + \frac 1 x = 1$

My favourite way of solving $x+\frac{1}{x}$ questions:

Let $x = e^{iy}$ ; then $\cos y = \frac{e^{iy}+e^{-iy}}{2} = \frac{1}{2}$ . One of the solutions for $y = \pi/3$ , but any other solution will work.

It follows that $x^{1729}+x^{-1729} = e^{1729iy}+e^{-1729iy} = 2\cos 1729y = 2\cos(1729\pi/3) = \boxed{1}$ .

If $x + \frac{1}{x} = 1$ , then $x^2 + 1 = x,$ or $x^2 - x + 1 = 0.$ Multiplying by $(x+1)$ , we see that $(x+1)(x^2-x+1) = 0,$ or by difference-of-cubes factorization, just $x^3 + 1 = 0.$ Thus, $x^3 = -1,$ or $x^6 = 1.$ Then, $x^{1729} = (x^6)^{288} \cdot x$ , so the expression just equals $x + \frac{1}{x}$ again, which is 1.

(x+1/x)=1.
x^2+(1/x)^2=-1.
x^4+(1/x)^4=1.
...... Look for periodicity. ...... x^1726+(1/x)^1726=-1.
x^1728+(1/x)^1728=1. ......eq.1.
@ Multiply eq.1 by x once(eq.2) and divide by x (eq.3).add eq.2and 3(eq.4).
T+x^1727+(1/x)^1727=1. (T =x^ 1729+(1/x)^1729).
Repeat the procedure in @.
Putting the known values, T=1.

If you have the same base and it's addition or subtraction you just need to add the exponent of the two base

1729+(-1729)=0

so the exponent is zero and we know that if the exponent is zero it is equal to one x to the power of 0= 1

The problem is very easy to solve. They have given , x + 1/x=1

=x + x^ -1= 1

=x^1 + x^ -1 =1 ......(1)

now if we follow the question, what we want is x ^1729 +x^-1729 to get 1729 and - 1729 as the power of bases of(1), I can multiply the power 1 and -1 with 1729 to get expected result: so even on the rhs, we multiply the power with 1729= 1^1729= 1.

I have not used the proper terms so plz excuse me.. whatever I have shown is the logic I used... I might be wrong, and I might be right, so please correct me if its the former case.

let f(n)=x^n+x^(-n); we can get: f(n)+f(n-2)=f(n-1); if f(1)=1; then f(2)=-1,f(3)=-2,f(4)=-1,f(5)=1,f(6)=2,f(7)=1,f(8)=-1; we can find that:f(m)=f(7m),and 1729/7=124,so f(1729)=f(1)=1

Given that x+1/x=1, So relating the above condition with x^1729+x^-1729 Gives 1 as the result.

Taking the 1729th root for the first equation will give the second equation then the result should be = 1 as the second equation

A very nice problem! As solution you can look at the therm x+1/x=1 and compare it to the expression x^1729+x^-1729. The expression x^-1729 equals to 1/(x^1729). Now I have excanged both x^1729 with x and got x+1/x, thats like in the therm equal to 1. In comparison to other solutions i don't know if my solution is the right way to solve this problem.

This does not only work for 1729, it works for all 6k + 1, where k is any integer.

Following Rajdeep Dhingra's solution, and replacing 1729 with 6k + 1, we have $S = 2\cos (\frac{(6k+1)\pi}{3}) = 2\cos(2k\pi + \frac{\pi}{3}) = 1$

The equation at the bottom means x = 0.5 Cancel out the 2 powers and you have 0.5+0.5 = 1

Nice solution

Let $a_{n} = x^n + x^{-n}$ . Then since $a_1 = 1$ by multiplying it out, $a_n = a_n.a_1 = a_{n+1} + a_{n-1}$ . This gives us $a_{n+1} = a_{n}- a_{n-1}$ .

Using this recurrence you see that the sequence of values has period 6, so $a_{1729} = a_1= 1$ .

ln x^y + ln 1/x^y= ln z

y ln x - y ln x=0=ln z

z=1

If we analize the series we get to know after finding two values (x+1/x=1) and (x^2 + 1/x^2= -1) Now the value of further series is difference of preciding two like (X^3 + 1/x^3 = -2) difference of values of (x^2 + 1/x^2) & (x^1 + x^1)

Doing the same for upto (x^6 + 1/x^6 = 2)

We now get that from x^7 the value will get repeat and it attains the value of same as of (x + 1/x) Which means the value x is changing as ( X^n + 1/x^n = X^(6k+m) )

Where n is the power of x can be written as 6k+m

Where k and m depends on the number i.e. x^13 K=2 & m=1

For m=1 it will attain the value of (x + 1/x =1) M=1. 1 M=2 -1 M=3 -2 M=4 -1 M=5. 1 M=6 2

now in given question N= 1729 Which can be written as 288*6+1 Here k= 288 and m= 1 So for m=1 value is 1

Just make the given equation as per the another one.

I know that many of you will laugh reading this but...

I simply solved it like that:

$x^{1729}+x^{-1729}$ with $x+\frac{1}{x}=1$

$x^{1729}+\frac{1}{x^{1729}}$

hence, [x=1]\

What's wrong in my reasonment? I'm a disaster at maths xD

Simply both x powers are combine -1729+1729=0 ,and x°=1.....

Since x+1/x= 1, then x^2 + 1 =x and 1+1/x^2 =1/x.

Then since (x+1/x)^2 = x^2+1/x^2+2, using identities derived above we further derive x^2+1/x^2+2 = x^2+1/x+1 = x + 1/x. Thus it also holds (x+1/x)^3 =1.

Since 1729 = 3^3 * 2^6+1.

Then It follows (x+1/x)^1729 = ((x+1/x)^3)^3 * ((x+1/x)^2)^6 *(x+1/x)^1 = 1, based on previous identities.

This for me was quite easy to solve without using too much work. All you have to know is that there cannot be negative powers, and if there are, you make them denominators. Therefore you would have

x^{1729} over x^{1729}

Simplify To get 1

Nice question given x+1/x=1 =x^2=x-1 Now x^4=(x-1)^2=x^2-2x+1=x-1-2x+1=-x

X^8=(x^4)^2=(-x)^2=x-1

X^9=x^8.x=(x-1).x=x^2-x=x-1-x=-1

X^81=(x^9)^9=(-1)^9=-1I

X^729=(x^81)^9=(-1)^9=-1

X^10=x^9.x=-x

X^100=(x^10)^10=(-x)^10=x^10=-x

X^1000=(x^100)^10=(-x)^10=x^10=-x

Now

X^1729+x^-1729 =x^729+1000+1/x^729+1000

=(x^729). (X^1000)+1/(x^729). (X^1000)

=(-1). (-x)+1/(-1). (-x)

=x+1/x

=1

Given expression x+1/x =1 shows that x can be expresses as 1^1729 or any other number in this equation ,therefore x^1729 + 1/x^1729 = 1 ans

First, we experiment with

$x+\frac{1}{x}$

Square it. Since

$(x+\frac{1}{x})^2 = x^2+\frac{1}{x^2} + 2\frac{x}{x} = 1$ ,

and we know that $1^2$ is $1$ so it goes on and the answer is 1