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What is the smallest perfect square whose last 3 digits are all the same digit?

After testing different numbers with a calculator, I have found that $38^2=1444$ is the smallest...how would you do so without using a calculator? I've tried writing as

$A \cdot 10^3 + 111 \cdot B = k^2$ and playing around with it, taking mod $3$ and $37$ (since $111=3\cdot37$), but those have yielded nothing...

Also, I know it must be 4 digit since $100^2=10,000$ is also a perfect square with its last 3 digits the same.

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    $\begingroup$ Once you know it is 4 digits, then since $\sqrt{1000}$ is about 33 so even if you have to multiply by hand, it won't take you long to find 38. $\endgroup$
    – Ted
    Commented Jul 29 at 23:56
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    $\begingroup$ You are trying to find the smallest so it makes sense to start with 33, 34, 35... $\endgroup$
    – Ted
    Commented Jul 30 at 0:07
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    $\begingroup$ Joke answer: $0^2=000$. $\endgroup$
    – Lucenaposition
    Commented Jul 30 at 0:09
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    $\begingroup$ Based on Mathematica experiments, I'd conjecture that 38 is in some sense the "only" nontrivial example. Specifically, any other example I've found is either of the form $n=100k$ (so $n^2$ ends in 000) or $n=\pm 38\mod{500}$. $\endgroup$
    – Semiclassical
    Commented Jul 30 at 0:25
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    $\begingroup$ @Semiclassical: according to my answer, the last three digits are $000$ or $444$, and the solutions to $n^2\equiv444\bmod1000$ are precisely $n\equiv\pm38\bmod500$ $\endgroup$
    – J. W. Tanner
    Commented Jul 30 at 13:23

3 Answers 3

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As pointed out by heropup in an answer that has now been deleted, the last digit of a square must be $0, 1, 4, 5, 6, $ or $9.$ But the last two digits of a square must leave remainder $0$ or $1$ when divided by $4$, so they can’t be $11$, $55$, $66$, or $99$. That leaves $000$ (which probably doesn’t count), $444$, $1000$, $\color{red}{1444}$…

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Assuming that $A = 0$, that is $k^2 = B\cdot111$, we arrive at a contradiction since $B \in \{0, \ldots, 9\}$ (and assuming $k$ is taken to be nonzero).

If $A = 1$, then $k^2 = 1 \mod 37$, so $(k+1)(k-1) = 0 \mod 37$. And because $37$ is prime, $k = \pm 1 \mod 37$. In other words $$ k = 37x \pm 1, $$ for some integer $x \in \{1, 2\}$ (as $k \leq 100$). The case $x = 1$ includes the correct value of $k = 38$.

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Let us note the 2-digit number $k = 10a + b$ and let's suppose that $k^2 = 100a^2 + 20ab + b^2$ has the last 3 digits being the same.

Then the last two digits must have the same parity, but the $100a^2 + 20ab$ part cannot change the parity of these digits, so actually $b^2$ must have $2$ digits of the same parity !

We can then eliminate $1^2 = 01$, $3^2 = 09$, $4^2 = 16$, $5^2 = 25$, $6^2 = 36$, $7^2 = 49$ and $9^2 = 81$ because the 2 digits have different parity.

This leaves only $b = 0$, $2$ or $8$ as possibilities, thus we have the repeating digit $B = 0$ or $4$.

If we start from $k = 32$ to have $k^2 \gt 1000$, we stumble pretty quick on the correct answer $k=38$.

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    $\begingroup$ What about $b=0$? $\endgroup$
    – J. W. Tanner
    Commented Jul 30 at 13:03
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    $\begingroup$ Ooops good point, adding the case $b = 0$ $\endgroup$
    – Vincent
    Commented Jul 30 at 13:29
  • $\begingroup$ I like your reasoning (+1), but I think you should replace the last paragraph with more of your reasoning: If $b=0$ then the smallest non-zero solution is $100^2$. If $b=2$ then the last two digits are $12$ or $62$. However $12^2=144$ so whatever the remaining digits, the "hundreds" digit will always be odd. If the last two digits are $62$ then the last three have to be $462$ or $962$, of which $462^2$ gives us the smallest answer. This leaves $b=8$. The last two digits are $38$ or $88$. However $88^2=7744$, whose third digit is odd, so no solutions will end $88$. $38^2= 1444$. $\endgroup$
    – tkf
    Commented Jul 30 at 16:25
  • $\begingroup$ So the smallest solution is $38^2$, $100^2$ or $462^2$, of which $38^2$ is clearly the smallest. $\endgroup$
    – tkf
    Commented Jul 30 at 16:27
  • $\begingroup$ In fact you get a complete list of three digit solutions: $38, 538, 462,962, 100,200,300,400,500,600,700,800,900$, which I see has already been enumerated [email protected]. $\endgroup$
    – tkf
    Commented Jul 30 at 16:36

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