I know that $10$..$01$ can’t be a perfect square, since it is $2$ mod $3$, but $10$..$09$ can a perfect square, since it is $1$ mod $3$.
Am I correct?
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1For some basic information about writing mathematics at this site see, e.g., basic help on mathjax notation, mathjax tutorial and quick reference, main meta site math tutorial and equation editing how-to. – Henrik supports the community Mar 29 '23 at 10:32
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1Just because a number is $1 \pmod 3$ it doesn't mean it can be a square, as the following example shows: $7$. – jjagmath Mar 29 '23 at 10:38
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Use the method(s) in the linked dupe (or hundreds of other dupes). – Bill Dubuque Mar 29 '23 at 18:46
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No it cannot. Perfect squares only have remainder in $\{ 0, 1, 3, 4, 5, 9 \}$ divided by $11$ (check it by yourself).
But $10^k + 9 \equiv 8 \text{ or } 10 \pmod{11}$.
Try different prime moduli aside from $3$ next time.
PinkRabbit
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Please strive not to post more (dupe) answers to dupes of FAQs, cf. recent site policy announcement here. – Bill Dubuque Mar 29 '23 at 18:43
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@BillDubuque "The essence of mathematics lies precisely in its freedom." Let us stop to ask for good questions. There are times in which people lost their way, or are really stuck on something. Freedom to ask poor (?) questions is a right. We are all here to learn, we are not here to demand quality. This is not a luxury site or something similar. Better a poor question denoting the will to know, that an ignorant "vote to close" the question et similia. – Heidegger Mar 15 '24 at 17:55