How can I solve $x^2 \equiv 19 \pmod {59}$?
I know that we can just try squaring numbers from 1 to 58 , but this is a very slow method, is not their a quicker one?
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What does your first line have to do with the question in the title? Is $(3/p)$ an attempt to write a Legendre symbol? – hmakholm left over Monica Jun 13 '19 at 13:27
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@Henning Makhlom I am sorry I edited the question – Emptymind Jun 13 '19 at 14:12
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@Intuition For further intuition see my answer - which shows that is it s specila case of an old result of Lagrange . – Bill Dubuque Jun 13 '19 at 14:27
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You can get it by playing around with equivalences modulo $59$ and extracting squares: from $$x^2 \equiv 19 \equiv -40 \equiv 2^2(-10) \pmod{59}$$ and $$-10 \equiv 49 \equiv 7^2 \pmod{59},$$ we have $x^2 \equiv 14^2 \pmod{59}$, so $x \equiv \pm 14 \pmod {59}$.
Théophile
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Lagrange observed that we can easily compute a square root in groups of odd order $\,\color{#c00}{2n+1},\,$ i.e. the equation $\,x^{\large 2} = a\,$ has solution $\,x = a^{\large n+1}\,$ by $\,x^{\large 2} = a\, a^{\large \color{#c00}{2n+1}} = a,\,$ by Lagrange's Theorem.
Here mod $\,\color{#0a0}{p = 4n+3}\,$ the subgroup $\,S\,$ of squares has odd order $\,(p-1)/2 = 2n+1.\,$ Thus, by above, any square $\,a\in S\,$ has a square root $\,x = a^{\large n+1}.\,$ Here $\,n\!+\!1 = \color{#0a0}{(p+1)/4} = 15$.
Bill Dubuque
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Is it similar to Shanks - Tonelli Algorithm for calculating modular square roots ? – Emptymind Jun 13 '19 at 22:34
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Why here $ n+1$ is given by your last line ? Why you divided $ p +1$ by 4? – Emptymind Jun 13 '19 at 22:43
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If I have mod 49 ... How can I calculate $ n+1 $ I calculated it and found that it is 13 but the final answer was wrong – Emptymind Jun 13 '19 at 23:15
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@Intuition $49$ is not a prime $,\color{#0a0}{p\equiv 3\pmod{4}},$ so the above does not apply.. – Bill Dubuque Jun 13 '19 at 23:21
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With $\phi(59)=58=2\cdot 29$ we know that all elements of the multiplicative group have order $1,2,29$ or $58$ (see the answer of Bill Dubuque for the named and more general order arguments). Test $$ 19^{29}=1\pmod{59}$$ computationally, giving the order of $19$ as the odd number $29$, which allows to compute a square root from $$19^{2⋅15}=19\pmod{59}.$$ This gives one (the?) pair of square roots as $$x=\pm19^{15}=\pm14\pmod{59}.$$
Lutz Lehmann
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@Aqua This is a special case of an old result of Lagrange - see my answer. – Bill Dubuque Jun 13 '19 at 14:26
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Is this what is called Shanks - Tonelli Algorithm for calculating Modular square roots? – Emptymind Jun 13 '19 at 22:30
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Hint: You Gauss reciprocity theorem:
$$\Big({19\over 59}\Big)\Big({59\over 19}\Big) = (-1)^{{59-1\over 2}\cdot {19-1\over 2}}= -1$$
Since $$59 \equiv 2\pmod{19}$$ and if $$x\equiv 0,\pm1,\pm2,\pm3,\pm4,\pm5,\pm6, \pm7,\pm8, \pm9 \pmod{19}$$ we have $$x^2\equiv 0,1,4,9,-3,6,-2, -8,6, 5 \pmod{19}$$
we see that $$\Big({59\over 19}\Big)=\Big({2\over 19}\Big) =-1$$
so $$\Big({19\over 59}\Big) =1$$ which means $19$ is a square modulo $59$
Now if you want explicit $x$ then you can use brute force method, if a modulo is not, like in this case, to big:
$$x≡0,±1,±2,±3,±4,±5,...±29(\mod59)$$ and then calculate the square of $x$ and we can see $14^2≡19(\mod59)$.
nonuser
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2But I want the values of $x$. I know that the given congruence has a solution. – Emptymind Jun 13 '19 at 13:18