$1^0 = 1\to 1 =1$

$x^1=x\to x=x\;\forall x$.

$9^2 = 81\to 8+1=9$

$8^3=512\to 5+1+2=8$.

$7^4=2401\to 2+4+0+1=7$

$46^5 = 205962976\to 2+0+5+9+6+2+9+7+6=46$

$64^6 = 68719476736\to 6+8+7+1+9+4+7+6+7+3+6 = 64$

$68^7= 6722988818432\to 6+7+2+2+9+8+8+8+1+8+4+3+2 = 68$

$54^8 = 72301961339136\to 7+2+3+0+1+9+6+1+3+3+9+1+3+6=54$

$71^9 = 45848500718449031$
$4+5+8+4+8+5+0+0+7+1+8+4+4+9+0+3+1 = 71$


Given a positive integer $b$, there exists a positive integer $a$ such that the digit sum of $a^b$ is equal to $a$.

Can this be proven? I don't know how to prove it; it was about 3:45am in the morning and I couldn't go to sleep, so I just went on my calculator and messed around because I was bored. That's when I noticed this cool property and decided to share it here.

It is now 4:35am so... I gotta go to bed. I'll see you in some hours and hopefully gather the time to work on this. Sorry about that!

Oh, incidentally, I also noticed that the digit sum of $29^5$ is $23$ and the digit sum of $23^5$ is $29$, so... there are cycles here. Same for $31$ and $34$. Also, the digit sum of $13^2$ is $16$ and the digit sum of $16^2$ is $13$. These cycles seem to only have two numbers involved, but I think regarding the seventh power, there is more than two involved (start with $72^7$ I think), however there is also a cycle between two numbers of seventh powers (between $44$ and $62$). Does this help? I don't know.

I have to go to bed. Good night!

Thank you in advance.

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    $\begingroup$ @Jakobian thanks for the edit! :) $\endgroup$ – Mr Pie Apr 19 at 18:39
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    $\begingroup$ @user477343 Go to bed! ;-) $\endgroup$ – amsmath Apr 19 at 18:45
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    $\begingroup$ @user477343 This looks like an interesting problem. However, as currently stated, the conjecture is always trivially true for $a = 1$. I assume you mean for $a \gt 1$. $\endgroup$ – John Omielan Apr 19 at 19:58
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    $\begingroup$ There is the trivial solution $1^b=1$, so perhaps the question needs to be stated more precisely. $\endgroup$ – Steven Clark Apr 19 at 20:32
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    $\begingroup$ For $b>25$ the possible $a$ are located in the range of $\{1,\ldots,b^2\}$. Reason: Let $s(n)$ be the decimal digit sum of $n$. Then, since the number of digits of $n$ is $1+\lfloor\log_{10}n\rfloor$, you have $s(n)\le 9(1+\lfloor\log_{10}n\rfloor)\le 9(1+\log_{10}n)$. Hence, $s(a^b)\le 9(1+b\log_{10}a)$. Put $f(a) := a-9(1+b\log_{10}a)$. Then $f'(tb^2) = 1-\tfrac 9{tb\log(10)} > 0$ for $t\ge 1$ and $b>25$. Moreover, $f(b^2) = b^2-18b\log_{10}b-9 > 0$ for $b>25$. This shows that $a\le 9(1+b\log_{10}a)$ can only happen if $a\le b^2$. $\endgroup$ – amsmath Apr 19 at 20:52

Sorry for having to put this in the answer section (as I don’t have the reputation to add comments yet) but this seems like a really interesting problem. I noticed with b=2, the value you get when you square it and add the digits seems to end up being 9 more frequently than average (the values I got were 1,4,9,7,7,9,13,10,9,1,4,9,16,16, 9 etc). This may be the first step. Alternatively you could plot the values of a and b, so plot (1,1), (2,4), (3,9), (4,7) etc and where these points fall onto the line y=x, you get a solution (for b=2, (9,9)).


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