I found pic related in a book about problems in number theory,

>>8669064
144

>>8669087
I know the answer. But I am wondering what they expect me to do to get it.

>>8669090
Galois Theory maybe ?

It's clear that
133 < n < 4^(1/5) * 133 < 176

The last digit of n can be found easily by considering the equation modulo 10.
n^5 = 3^5 + 0^5 + 4^5 + 7^5 (mod 10)
= 4 (mod 10)
Which means that n^5 and therefore n has 4 as the last digit.

So you only have to test n=134, 144, 154, 164, 174.

It turns out that n=144 works.

>>8669090
But it takes like 0.01sec to "bruteforce" (it's a simple exponentiation algorithm which is standard in any language today).
I'd just type that in python if i'd really have to solve it. Otherwise I'd say quintic function.

>>8669095
The book is in 「Elementary」Number Theory. I doubt that they would make you use Galois Theory.

>>8669100
>But it takes like 0.01sec to "bruteforce"

It is meant to be a pen and paper problem. The objective is to not bruteforce it. Like >>8669099 does.

I simply did not know how.

>>8669099
>133 < n < 4^(1/5) * 133 < 176

How do you know this?

Look at the equation modulo 2:
n^5 = 0 (mod 2)
n = 0 (mod 2)
Look at the equation modulo 3:
n^5 = 0 (mod 3)
n = 0 (mod 3)
Look at the equation modulo 5:
n^5 = 4 (mod 5)
n = 4 (mod 5)
Look at the equation modulo 7:
n^5 = 2 (mod 7)
n = 4 (mod 7)

The Chinese Reminder theorem says that
n = 144 (mod 210)
for an integer k.

So n must be 144.

>>8669111
>How do you know this?
133^5 < 133^5 + 110^5 + 84^5 + 27^5 < 4 * 133^5

>>8669064
n needs to be larger than 133, and be congruent to 4 modulo 5.