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Consider the quadratic congruence \begin{equation} x^2-34x+1 \equiv 0 \pmod{p}, \end{equation} where $p \equiv 7 \pmod{8}$. There are two solutions, since we can write it as \begin{equation} (x-17)^2 \equiv 288 \pmod{p} \end{equation} and check that $288$ is a quadratic residue in this case. Since the product of solutions is congruent to $1 \pmod{p}$, then both of them are either quadratic residues or quadratic non-residues. I have checked multiple primes, and it appears that both solutions are always quadratic residues. I'm not sure how to prove it; I feel like I'm missing something obvious. Any hint would be much appreciated!

J. W. Tanner
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J. Miarecki
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2 Answers2

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Easy way: $ $ applying my denesting formula we can compute $\color{#0a0}{\sqrt x}\,$ by completing a square

$\qquad \qquad\ 34x = x^2+1 \iff 36x = (x\!+\!1)^2\!\!\iff x = \left(\!\color{#0a0}{\dfrac{x\!+\!1}6}\!\right)^{\!2}$

Remark $ $ So $\:\!x\:\!$ has a $\:\!\rm\color{#0a0}{square\ root}\:\!$ in any ring where $\color{#0a0}{1/6}\,$ exists, e.g. $\:\!\Bbb Z_n\:\!$ for all $\:\!n\:\!$ coprime to $\:\!6,\,$ e.g. OP's $\,n=$ prime $p\equiv 7\pmod{\!8},\,$ where $\,p\neq 2,3\Rightarrow p\,$ coprime to $\:\!6$.

Bill Dubuque
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  • It should be noted that $x$ does not exist in all rings, e.g., $\mathbb Z_5$ and $\mathbb Z_{11}$ – J. W. Tanner Mar 15 '24 at 17:59
  • @JWT Of course, the proof shows that if a root $x$ exists then $x = ((x+1)/6)^2,,$ assuming $1/6$ exists. But it is given in the OP that a root exists (so we don't need to prove it) – Bill Dubuque Mar 15 '24 at 18:14
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When $p\equiv7\pmod8$, then $y^2=2$ has solutions $\pmod p$,

which I will write as $y=\pm\sqrt2\pmod p$.

The solutions of the congruence $x^2-34x+1\equiv0\pmod p$

are seen to be $x=17\pm12\sqrt2\pmod p,$ using the quadratic formula.

Can you show that $x=(3\pm2\sqrt2)^2$?

J. W. Tanner
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    You should give the details of how you computed $\sqrt{17\pm 12\sqrt 2},,$ rather than pull it out of a hat like magic. But there is need to use QR or the quadratic formula - see my answer. – Bill Dubuque Mar 15 '24 at 00:40
  • You are correct. I could have bypassed the quadratic formula and said $(x-17)^2\equiv288\iff $ $x\equiv17\pm12\sqrt2$. And I recognized $17\pm12\sqrt2$ as a square from the units in $\mathbb Z[\sqrt2],$ related to the $n=2$ case of Pell's equation, but you're right that denesting is the obvious approach when that isn't recognized – J. W. Tanner Mar 15 '24 at 02:28
  • To clarify, my point is that it is much simpler (high-school algebra!) to simply apply the equation-form of the denesting formula as in my answer, i.e. completing the square shows $,x = ((x+1)/6)^2$ is a square, QED. We don't need to use deeper quadratic reciprocity as you do, nor do we need to actually compute the roots, then use (unexplained!) deeper Pell theory to take their square root. Do you really think the OP knows how to apply said Pell theory? You should explain it if you use it. – Bill Dubuque Mar 15 '24 at 17:39