Numbers That Reappear in Their Squares

A three-digit number has this odd habit: the last three digits of its square are the number itself, in the same order. Find it. How many three-digit numbers behave this way?

Solution

There are two: $376$ and $625$, since $$376^2 = 141\,376, \qquad 625^2 = 390\,625.$$ To see why these and no others, we want a three-digit $n$ with $n^2$ ending in $n$, that is $n^2 - n = n(n-1)$ divisible by $1000 = 8 \times 125$. Now $n$ and $n - 1$ are consecutive, so they share no common factor; whatever powers of $2$ and $5$ make up the $1000$ cannot be split between them, so the whole $8$ goes into one of the two and the whole $125$ into the other. That gives two interesting cases:

• $8$ divides $n$ while $125$ divides $n - 1$. So $n$ is a multiple of $8$ that is one more than a multiple of $125$, and the only such number below $1000$ is $n = 376$ (indeed $376 = 8 \times 47$ and $375 = 125 \times 3$).

• $125$ divides $n$ while $8$ divides $n - 1$. So $n$ is a multiple of $125$ that is one more than a multiple of $8$, which gives $n = 625$ (since $625 = 125 \times 5$ and $624 = 8 \times 78$).

The only other possibilities, where $1000$ divides $n$ or $n - 1$ outright, give numbers ending in $000$ or $001$, not a genuine three-digit answer. So $376$ and $625$ are the only ones. Such numbers are called automorphic. Pleasingly $376 + 625 = 1001$, and the habit persists into more digits: $625^2 = 390625$ ends in $0625$, and $9376^2 = 87909376$.