# A special square

Is there a four-digit square number which has at least one digit in common with every other four-digit square?

• There's only one four-digit perfect number (8128) and it isn't a square. There probably aren't all that many people who are bothered by this unfortunate coincidence with the particular choice of words, so I'll leave it up to you to decide whether to remove the word "perfect" from the question. – Bass Jun 28 '18 at 7:09
• @Bass Square numbers are sometimes known as "perfect squares" (and bear no relation to perfect numbers) – phenomist Jun 28 '18 at 9:23

Yes. $6241 = 79^2$ is the only number that satisfies this condition.
Suppose that our square number doesn't contain the digit 0. Then it must share at least one digit in common with $2500 = 50^2$, $3600 = 60^2$, $4900 = 70^2$, and $8100 = 90^2$, which each have distinct pairs of digits from the others. So it must contain a digit from each of these pairs, and can't contain the digit 7. Since it must also share a digit with $7744 = 88^2$, it must contain the digit 4 and not 9. Now consider $2209 = 47^2$, forcing the digit 2 and not 5. Next, $7056 = 84^2$ to force the 6 and not the 3. From here, I looked at all of the possible square numbers and found that all of them contained a 2, 4, 6, and either 1 or 8. Moreover, there is just one square that contains all of these, and that is $6241 = 79^2$.
Now suppose that our square number does contain the digit 0. Next, let's check whether it contains the digit 4. If it doesn't, then it contains both a 1 and a 7, from $1444 = 38^2$ and $7744 = 88^2$. But a scan through the list shows that the only squares with both a 0 and a 7 are $2704 = 52^2$ and $7056 = 84^2$, neither which contain a 1. So it must contain the digit 4.
It must share a digit with $6889 = 83^2$, so that eliminates a bunch of possibilities: {4096, 4900, 6084, 6400, 9409, 9604}
Next it must share a digit with $1225 = 35^2$, which eliminates everything else. So there are no further solutions.