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I am not familiar with the topic. I can work out for the problem when the problem has a solution. But here I don't know how to it. Using the table I can see there is no solution. But I want to know how to do mathematically.

Thomas Andrews
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Madhan Kumar
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    Compute $x^2 \textrm{ mod } 11$ for $x=0,1,\dots,10$. I think this approach is "mathematical". – azif00 May 10 '21 at 20:51
  • I think by, "I want to know how to do mathematically", you really mean, "I want to know how to do it without using the table"... – Adam Rubinson May 10 '21 at 20:52
  • Checking all 11 possible values of $x$ is a perfectly mathematical way of doing this. – lisyarus May 10 '21 at 20:52
  • @azif00 You can save one $x$ or six $x$'s. – WhatsUp May 10 '21 at 20:52
  • No other way??? – Madhan Kumar May 10 '21 at 20:52
  • One "other way" is to say that $11 \equiv 3\mod 8$. Another "other way" is to say that $2 \equiv -3^2 \mod 11$ and $11 \equiv 3 \mod 4$. – WhatsUp May 10 '21 at 20:53
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    There is a general theorem that, for odd primes $p,$ $x^2\equiv 2\pmod p$ has a solution if and only if $p\equiv \pm 1\pmod 8.$ – Thomas Andrews May 10 '21 at 20:53
  • @lisyarus, it only works with small modulus – PNT May 10 '21 at 20:54
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    @Yassir it "works" with any modulus, it's just not practical. (and, well, 11 is small enough) – lisyarus May 10 '21 at 20:55
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    @MadhanKumar you could only check $x^2\pmod{11}$ for $x=0,1,\dots,5$, since $x^2\equiv(-x)^2\pmod{11}$. – Kenta S May 10 '21 at 20:56
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    In general, for prime $p,$ and any $n$ not divisible by $p$ $x^2\equiv n\pmod p$ has a solution if and only if $n^{(p-1)/2}\equiv 1\pmod p.$ Here, $n=2,p=11,$ then $2^5=32\not\equiv 1\pmod{11}.$ – Thomas Andrews May 10 '21 at 20:56
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    If you want you can try "my beginner method". $x=11k±1,±2,±3,±4±5$ it doesn't require more time. At most 2 minutes. You can prove. – lone student May 10 '21 at 20:58
  • Apply Euler's Criterion, i.e. raise $,x^2\equiv 2,$ to power $, (p-1)/2 = 5,$ to get $,x^{10}\equiv -1,,$ contra lil Fermat, as in this answer in the 2nd dupe. – Bill Dubuque May 10 '21 at 21:48
  • I think it's better and "more mathematical" to hit a problem with a rook a bunch of times then to go to the deep end of the pool, borrow a magic formula you don't understand and try to apply it. Work your way to understanding the method Thomas Andrews and Bill Dubuque suggest (they are both excellent suggestions) but play with taking the $11$ possible classes of $x$ ans squaring them and seeing what you get until you start feeling things. – fleablood May 10 '21 at 22:34

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