21
$\begingroup$

Prove that the number $14641$ is the fourth power of an integer in any base greater than $6$?

I understand how to work it out, because I think you do

$$14641\ (\text{base }a > 6) = a^4+4a^3+6a^2+4a+1= (a+1)^4$$

But I can't understand why they had to specify that the base is greater than 6? Is that because if it's 4, then 4 will be cancelled out in the equation and if it's 6, it will be too?

Please advise.

Sorry for asking such a trivial question.

$\endgroup$
  • 6
    $\begingroup$ Similarly, the number 15AA51 is a 5th power of an integer for any base greater than 10 (A is the hexidecimal representation of 10). If base 5, $6^4 = 5^4 + 4*5^3 + 6*5^2 + 4*5 + 1 = 10_5^4 + 4*10_5^3 + 11_5*10_5^2 + 4*10_5 + 1 = 10_5^4 + 10_5*10_5^3 + 10_5^2 + 4*10_5 + 1 = 2*10_5^4 + 0*10_5^3 + 10_5^2 + 4*10_5 + 1 = 20141_5$. $\endgroup$ – fleablood Jan 30 '16 at 21:44
32
$\begingroup$

In base $6$, $14641$ is not a valid representation of a number, because $6$ is not a valid digit in base $6$.

It is true that $1+4\cdot 6 + 6\cdot 6^2+4\cdot 6^3+6^4$ is still equal to $7^4$. But the proper representation for this number in base $6$ is $15041_6$.

$\endgroup$
10
$\begingroup$

Your question generalizes. The expansion of $(b+1)^n$ is clearly a perfect $n$th power for every base $b$ greater than the largest binomial coefficient (the middle one). When $n > 4$ the middle binomial coefficient has more than one (decimal) digit, so you can't write the expansion using the digits $0, \ldots 9$.

Even more generally, the expansion of $(ab+c)^n$ is a perfect $n$th power when the base $b$ is greater than the maximum of the terms ${n \choose i}a^{i}c^{n-i}$. That's the condition that there be no carries when you calculate using the standard (in the U.S.) multiplication algorithm in base $b$.

Unfortunately, you can't get far when you're limited to base 10 digits. The two digit number powers you can compute before you have to start carrying are $11^4 = 14641$, $12^2 = 144$, $13^2 = 169$, $22^2 = 484$ and the reversals $21^2 = 441$ and $31^2 = 961$.

More examples:

$(110 + 1)^3 = 111^3 = 1367631$, $112^2 = 12544$, ...

$1111^2 = 1234321$, ... , $(11\ldots1)^2 = 12345678987654321$

For homework: find them all.

$\endgroup$
4
$\begingroup$

As a matter of personal preference, I like $b$ to represent the base rather than $a$.

So then $(b + 1)^2 = b^2 + 2b + 1$. And $(b^2 + 2b + 1)^2 = b^4 + 4b^3 + 6b^2 + 4b + 1$.

These facts are true whether $b$ is an ordinary positive integer greater than 1 or a more "exotic" number, like $\sqrt{-2}$. But whether $(b + 1)^4$ gets represented as 14641 in base $b$, that's a slightly different story.

Consider for example $b = 2$. Indeed $3^4 = 2^4 + 4 \times 2^3 + 6 \times 2^2 + 4 \times 2 + 1$. But the problem is that here it turns out that $6 > b^2$ and $4 > b$. Then $4b^3$ requires more than four binary digits to represent, $6b^2$ requires more than three bits and $4b$ requires more than two bits. So the binary representation of 81 is 1010001 rather than 14641.

Some of these problems persist through $b = 6$, because although $4b^3 < b^4$, we still have $6b^3 > b^3$. Then 2401 is 15041 in base 6.

The smallest integer $b$ satisfying $4b^3 < b^4$, $6b^2 < b^3$ and $4b < b^2$ is $b = 7$. Indeed 4096 in base 7 is 14641.

$\endgroup$
  • 1
    $\begingroup$ I think the use of b for the base is more common $\endgroup$ – phuclv Jan 31 '16 at 8:26

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.