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Lucky numbers are 4 digit numbers that have the following property: they are equal to the sum of the fourth power of their digits. Therefore, they can be expressed as follows:

$$1000a+100b+10c+d = a^4+b^4+c^4+d^4$$

What are all the lucky numbers? Please don’t use computers.

Hints:

Use Euler’s theorem.

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Observe that every 4th power is

congruent to one of $0,1,5,6$ modulo 10. This is because $1,3,7,9$ go to $1$ by Euler's theorem, $2,4,6,8$ go to $6$ since $2^4=16$ and $6^2=36$, and $5$ goes to $5$ and $0$ goes to $0$. This places a lot of restrictions on the sum, and knowing $d$ significantly reduces the possibilities for the set $\{a,b,c,d\}$.


If one of $a,b,c,d$ is $9$, then

all the others must be at most $7$ (otherwise the RHS is too big), and $a$ must be at least $6$ (from the LHS) therefore at least $7$ (from the RHS).

  • If there is a $7$, then we must have $a=9$, so the other two digits are at most $4$ (from the RHS). Here there is

    one solution, namely $9474=9^4+4^4+7^4+4^4$.

  • If there is no $7$, then we must have either $a=6$ or $a=9$; since $6^4+9^4>7000$, it must be $a=9$. The LHS is over 9000, so the other three numbers must be $6,6,?$ or $6,5,5$.

    There are no possibilities here.


In the remaining cases, none of $a,b,c,d$ can be $9$. If two of them are $8$, then

$a=8$ and the other two numbers must be at most $5$.

  • If there is a $5$, then $8^4+8^4+5^4>8800$ so $a=b=8$.

    No possibilities here.

  • If there is no $5$, then $8^4+8^4+4^4+4^4<8800$, so $b\neq8$, so $b\leq4$. Quickly checking possibilities, we find that

    having a $4$ isn't workable, nor is having a $3$, and indeed the only possibility here is $8208=8^4+2^4+0^4+8^4$.


Now we're left with smaller possibilities: none of $a,b,c,d$ is $9$ and at most one of them is $8$.

Modulo $3$, we know

$x^4\equiv1$ by Fermat's little theorem, so $a+b+c+d\equiv1$ if all of $a,b,c,d$ are coprime with $3$ or if three of them are multiples of $3$, $\equiv0$ if one of them or all of them are multiples of $3$, $\equiv2$ if two of them are multiples of $3$.

  • If three or four of them are multiples of $3$, then

    there are just three choices for each of the four digits: $\{0,3,6\},\{0,3,6\},\{0,3,6\},\{1,4,7\}$ if three are multiples of $3$, $\{0,3,6\},\{0,3,6\},\{0,3,6\},\{0,3,6\}$ if all four are, and these can be eliminated by hand.

  • If exactly one of them is a multiple of $3$, then

    the other three sum to a multiple of $3$, so they must be all the same modulo $3$. We have one of $\{0,3,6\}$ and either three of $\{1,4,7\}$ or three of $\{2,5,8\}$. Again these can be eliminated by hand.

  • If none of them are multiples of $3$, then

    one of them is congruent to $1$ mod $3$ and the other three sum to a multiple of $3$ so they must be all the same modulo $3$. We have one of $\{1,4,7\}$ and either three more of $\{1,4,7\}$ or three of $\{2,5,8\}$. Again these can be eliminated by hand.

  • Finally we have the case where two of the digits are multiples of $3$ and two aren't, so

    we must have two of $\{0,3,6\}$ and two of $\{1,4,7\}$. Playing around with the possibilities, we find that the only option here is $1634=1^4+6^4+3^4+4^4$.


Overall then, the lucky numbers are

$1634,8208,9474$.

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  • $\begingroup$ Beautiful explanation! $\endgroup$ – shanylong Aug 4 at 7:41
  • $\begingroup$ Hi. how did you format proper MathJax in your spoilers? I have tried to do it with $ but failed. $\endgroup$ – aminabzz Aug 6 at 21:01
  • $\begingroup$ @aminabzz MathJax should work in exactly the same way inside a spoilertag as outside. What sometimes messes up inside spoilertags is linebreaks - maybe that's what caused your failure? You can click the edit button on this post to see how the formatting is done. $\endgroup$ – Rand al'Thor Aug 6 at 21:02

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