# Number whose Square and Cube use all Digits Once

## Theorem

The only integer whose square and cube use each of the digits from $0$ to $9$ exactly once each is $69$.

## Theorem

First the limits are established for the square and cube of an integer to have exactly $10$ digits between them.

 $\displaystyle 46^2$ $=$ $\displaystyle 2116$ $\displaystyle 46^3$ $=$ $\displaystyle 97 \, 336$ $\displaystyle 47^2$ $=$ $\displaystyle 2209$ $\displaystyle 47^3$ $=$ $\displaystyle 103 \, 823$

 $\displaystyle 99^2$ $=$ $\displaystyle 9801$ $\displaystyle 99^3$ $=$ $\displaystyle 970 \, 299$ $\displaystyle 100^2$ $=$ $\displaystyle 10 \, 000$ $\displaystyle 100^3$ $=$ $\displaystyle 1 \, 000 \, 000$

So integers smaller than $47$ do not have enough digits available, and integers greater than $99$ have too many.

$47$ and $99$ themselves are eliminated on inspection of the above.

It is noted that integers ending in $0$, $1$, $5$ and $6$ have squares and cubes ending in those same digits.

Such numbers can be eliminated from our search, as they will duplicate the appearance of those digits.

It remains to check the integers between $48$ and $98$.

First the squares which duplicate at least one digit are eliminated:

 $\displaystyle 58^2$ $=$ $\displaystyle 3364$ which duplicates $3$ $\displaystyle 62^2$ $=$ $\displaystyle 3844$ which duplicates $4$ $\displaystyle 63^2$ $=$ $\displaystyle 3969$ which duplicates $9$ $\displaystyle 67^2$ $=$ $\displaystyle 4489$ which duplicates $4$ $\displaystyle 68^2$ $=$ $\displaystyle 4624$ which duplicates $4$ $\displaystyle 77^2$ $=$ $\displaystyle 5929$ which duplicates $9$ $\displaystyle 83^2$ $=$ $\displaystyle 6889$ which duplicates $8$ $\displaystyle 88^2$ $=$ $\displaystyle 7744$ which duplicates $4$ and $7$ $\displaystyle 92^2$ $=$ $\displaystyle 8464$ which duplicates $4$ $\displaystyle 94^2$ $=$ $\displaystyle 8836$ which duplicates $8$ $\displaystyle 97^2$ $=$ $\displaystyle 9409$ which duplicates $9$

Of the remainder, the cubes which duplicate at least one digit are eliminated:

 $\displaystyle 48^3$ $=$ $\displaystyle 110 \, 592$ which duplicates $1$ $\displaystyle 49^3$ $=$ $\displaystyle 117 \, 649$ which duplicates $1$ $\displaystyle 52^3$ $=$ $\displaystyle 140 \, 608$ which duplicates $0$ $\displaystyle 53^3$ $=$ $\displaystyle 148 \, 877$ which duplicates $7$ and $8$ $\displaystyle 54^3$ $=$ $\displaystyle 157 \, 464$ which duplicates $4$ $\displaystyle 57^3$ $=$ $\displaystyle 185 \, 193$ which duplicates $1$ $\displaystyle 64^3$ $=$ $\displaystyle 262 \, 144$ which duplicates $2$ and $4$ $\displaystyle 72^3$ $=$ $\displaystyle 373 \, 248$ which duplicates $3$ $\displaystyle 74^3$ $=$ $\displaystyle 405 \, 224$ which duplicates $2$ and $4$ $\displaystyle 78^3$ $=$ $\displaystyle 474 \, 552$ which duplicates $4$ and $5$ $\displaystyle 79^3$ $=$ $\displaystyle 493 \, 039$ which duplicates $3$ and $9$ $\displaystyle 82^3$ $=$ $\displaystyle 551 \, 368$ which duplicates $5$ $\displaystyle 87^3$ $=$ $\displaystyle 658 \, 503$ which duplicates $5$ $\displaystyle 89^3$ $=$ $\displaystyle 704 \, 969$ which duplicates $9$ $\displaystyle 98^3$ $=$ $\displaystyle 941 \, 192$ which duplicates $1$ and $9$

Of the remainder, the squares and cubes which duplicate at least one digit between them are eliminated:

 $\displaystyle 59^2$ $=$ $\displaystyle 3481$ which duplicate $3$ between them $\displaystyle 59^3$ $=$ $\displaystyle 205 \, 379$

 $\displaystyle 73^2$ $=$ $\displaystyle 5329$ which duplicate $3$ and $9$ between them $\displaystyle 73^3$ $=$ $\displaystyle 389 \, 017$

 $\displaystyle 84^2$ $=$ $\displaystyle 7056$ which duplicate $0$, $5$ and $7$ between them $\displaystyle 84^3$ $=$ $\displaystyle 592 \, 704$

 $\displaystyle 93^2$ $=$ $\displaystyle 8649$ which duplicate $4$ and $8$ between them $\displaystyle 93^3$ $=$ $\displaystyle 804 \, 357$

Finally:

 $\displaystyle 69^2$ $=$ $\displaystyle 4761$ $\displaystyle 69^3$ $=$ $\displaystyle 328 \, 509$

The result follows.

$\blacksquare$