Tribute to Ramanujan

Algebra Level 3

${ x }^{ 1729 }={ 1729 }^{ x }$

Find the integral root of the above equation.

The answer is 1729.

2 solutions

Swapnil Das
Sep 5, 2015

${ x }^{ 1729 }={ 1729 }^{ x }$

Taking logarithms on both sides, $x \log 1729 = 1729 \log x$

$\frac { \log { x } }{ x } = \frac { \log { \ 1729 } }{ 1729 }$

$x = 1729$

You have just found a solution of 1729. You need to prove that it's the only solution.

Pi Han Goh - 5 years, 9 months ago

Hmm… I am trying to do so.

Swapnil Das - 5 years, 9 months ago

Just differentiate $\frac{\log(x)}{x}$ . ;)

Maggie Miller - 5 years, 9 months ago

@Maggie Miller – Why we need to differentiate ?

If we differentiate we will get ::

{ x }^{ -2 }(-\log { (x) } +1)\quad =\quad 0\ \log { (x) } \quad =\quad 1

Syed Baqir - 5 years, 9 months ago

@Syed Baqir – Ah no, I mean differentiating $\log(x)/x$ shows that $\log(x)/x$ is decreasing for $x>e$ . Then $1729$ is the only answer greater than $2$ , and its easy to rule out $1,2,0$ and negative integers.

Maggie Miller - 5 years, 9 months ago

@Maggie Miller – Hmm, Now I know what you mean.

Becaues the derivative gives x = e ( I think critical point ),

Syed Baqir - 5 years, 9 months ago

@Bill Bell Inspired by you!

Swapnil Das - 5 years, 9 months ago

Didn't think of that!!!

Yuki Kuriyama - 5 years, 9 months ago

Haha, no problem :)

Swapnil Das - 5 years, 9 months ago

Syed Baqir
Sep 5, 2015

If L.H.S = R.H.S

Then x = 1729 < From Bases

1729 = x <From Exponents

1729 is a square free number !!

This argument would work if $1729$ were prime, but unfortunately it isn't. As a counterexample to the logic, by your argument the only integral solution to $x^4=4^x$ is $4$ - but this is false, since $2$ is also a solution.

However, I think you can easily fix your solution if you use the fact that $1729$ is square-free! :)

Maggie Miller - 5 years, 9 months ago

Thanks, I will update my solution.

Syed Baqir - 5 years, 9 months ago

Tribute to Ramanujan

The answer is 1729.

2 solutions

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