Reverse power a number

Question

So let's say I have a number, $n$, which is $327429$. I want to find which power of a number will result in this number. For example, I have the equation: $6^4$. This results in $1296$. Now what I want to do, is to find which power will result in my number, $n$ - let's call the base $b$ and the exponent $e$. So my solution to the problem would look something like $n = b^e$.

The base and exponent can be anything, as long as they are not decimals - they need to be integers with no decimal places. I'm planning to implement this in C++ (though I'm sure this would not really qualify as a coding question.)

Restating the Problem: I need to find $b$ and $e$ and I already know $n$.

How would I go about doing this?

Power detection is a special case of the linked post (see the algorithm of Bernstein linked there). — Bill Dubuque, Mar 28 '22 at 00:00
The answer you accepted using prime factorization is only usable for very small numbers because factorization is hard for larger numbers. But the algorithm I linked above doesn't suffer from this problem (it's essentially linear time). So your acceptance of that answer may mislead future readers. — Bill Dubuque, Mar 28 '22 at 01:30

score 1 · Answer 1 · answered Mar 27 '22 at 23:31

1

Find factorization $n = p_1^{a_1} \cdot p_2^{a_2} \cdots p_n^{a_n}$. Find greatest common divisor of $a_1, \ldots, a_n$, let's say $k$ divides all of these, so that $a_i = k \cdot u_i$. Then, if $k > 1$, we can write $n = b^e$, where $b = p_1^{u_1} \cdots p_n^{u_n}$, and $e = k$. If $k = 1$, such expression $n = b^e$ does not exist.

answered Mar 27 '22 at 23:31

xyzzyz

7,884

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Well, but factoring isn't known to be easy though, in that there is no fast algorithm [poly in number of digits in $n$] that factors an integer $n$. This approach yields a slow algorithm, whereas [as in the answer below] checking whether a number is a perfect $e$-th power for any fixed $e$ can be done quickly, and the only integers $e$ that need to be checked are the positive integers no larger than $\log_2 n$ which in turn is no more than $4×$ the number of digits of $n$. – Mike Mar 27 '22 at 23:48
Please strive not to add more (dupe) answers to dupes of FAQs, cf. recent site policy announcement here. – Bill Dubuque Mar 27 '22 at 23:57
Please don't bother me with requests like that after you marked it as a dupe. I didn't know it's a dupe. – xyzzyz Mar 28 '22 at 00:02

score 1 · Accepted Answer · answered Mar 27 '22 at 23:34

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If we want $e>1$ such that $n = b^e$, then $e$ can only be one of the numbers $2, 3, \dots, \lfloor \log_2 n\rfloor$, which is a very tractable number of cases. For each value of $e$, just compute the real number $n^{1/e}$, round it to get a candidate value of $b$, and then see if $b^e$ is actually equal to $n$. (Stop once it's clear that $n^{1/e} < 2$; then even $2^e$ is bigger than $n$, so $b^e$ will certainly be bigger than $n$ for any possible $b$.)

Number-theoretic approaches like trying to factor $n$ are going to be much slower, since factoring is hard.

answered Mar 27 '22 at 23:34

Misha Lavrov

159,700

1

Please strive not to add more (dupe) answers to dupes of FAQs, cf. recent site policy announcement here. – Bill Dubuque Mar 27 '22 at 23:57
My motivation here was primarily to avoid the main answer here being an incorrect answer. I think that this is more important than avoiding duplicate answers. – Misha Lavrov Mar 28 '22 at 01:19
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You could have simple left a comment on the answer - which would make the critique much more evident (even the OP seems to have missed it). – Bill Dubuque Mar 28 '22 at 01:30
@BillDubuque Gotcha, thanks. – HackerDaGreat57 Mar 28 '22 at 19:10

Reverse power a number

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