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I am asked to find $x\in\mathbb{Z}$ s.t. $|x^2+1|_5\leq5^{-4}$.

I have a method of doing it but I'm not sure if it's right and whether my conclusion is correct either. It also seems exceedingly long and not very efficient so perhaps there might be a better method of going about it.

Here's what I did:

First let $x_0=a_0=2$ and this gives $|x_1^2+1|_5=5^{-1}$.

Next consider $x_1^2=(a_0+5a_1)^2=-1\mod5^2$, for which I solve to get $a_1=1$. This in turn gives me that $|x_1^2+1|_5=5^{-2}$, so I think I may be on the right track. I then considered $x_2^2=(a_0+5a_1+5^2a_2^2)^2=-1\mod5^3$ but then after doing all the algebra, I got $4a_2^2=3 \mod 5$ which has no solutions since 3 is not a quadratic residue modulo $5$. Apart from the initial choice of $a_0$, there is no room for choices for the rest of the $a_i$'s.

The other option was for me to let $x_0=a_0=3$. This gives me $a_1=3$ but when I worked out the algebra for $a_2$, I got $a_2^2=2 \mod 5$, which again has no solution for the same reason.

Is this sufficient to say that no such $x$ exists for my equation?

Michael Hardy
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Haikal Yeo
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    Why do you use $a_2^2$ ? Can't you just call it $a_2$ ? – zozoens Jan 30 '14 at 14:09
  • Yeah. No idea why I did that! Just wondering, because $a_0$ has 2 choices, does that mean that there are 2 possible solutions? – Haikal Yeo Jan 30 '14 at 14:12
  • Yes. I don't know how much you know about $p$-adic numbers, but the numbers you are computing are approximations of the solutions of the equation $x^2+1=0$ in $\mathbb Q_5$ which is a field, so this equation can have at most $2$ solutions. The fact that they exist is what is called Hensel's lemma (see http://en.wikipedia.org/wiki/Hensel%27s_lemma). – zozoens Jan 30 '14 at 14:17
  • Yup I know about Hensel's Lemma. Is there also a more efficient way of doing this or is it literally just solving for $a_i$'s in the manner to get the answer? I realised that it does get more tedious when working in $\mod 5^3$ – Haikal Yeo Jan 30 '14 at 14:23
  • Yes there will be two solutions. Effectively you are approximating $\sqrt{-1}$. And like all square roots (in a field of characteristic different from two) differ in sign. – Jyrki Lahtonen Jan 30 '14 at 14:23
  • I happen to use a 5-adic square root of $-1$ as an example in this answer. Use at your own peril :-) – Jyrki Lahtonen Jan 30 '14 at 14:25
  • Basically, you always work in $\mod 5$: since you find your previous coefficients, you can always simplify by a factor of $5^{n-1}$, so you have an equation of degree $1$ on $\mathbb Z_5$ to find each of the coefficients. – zozoens Jan 30 '14 at 14:25
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    @Haikal: You may have made things a bit more complicated than they need to be by writing them out digitwise. e.g. you could have just written $x_1 = 2 + 5 a_1$ and $x_2 = 7 + 5^2 a_2$. If you need a lot of digits, you can switch over to Newton's method: or equivalently, note that if $$x^2 + 1 \equiv 0 \pmod{5^k}$$ then $$(x + 5^k y)^2 + 1 \equiv x^2 + 2 \cdot 5^k x y + 1 \pmod{5^{2k}}$$ –  Jan 30 '14 at 17:06

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You are working in the ring $\mathbb Z/(5^4\mathbb Z)$, or if you’re very well-educated, in the $5$-adic integers $\mathbb Z_5$. Either way, your quickest route to a root of a polynomial is Newton-Raphson, which I hope you know from high-school Calculus. The function whose root you want is $f(x)=x^2+1$, and its derivative is $f'(x)=2x$. If your first guess is $x_0=2$ (since $2^2+1\equiv0\pmod5$), you calculate the error, which is $5$, and divide by $f'(2)=4$ and subtract this from $2$ to get $3/4=x_1$, your second guess. Now the error is $25/16$, and you divide by $2x_1$ and subtract from $x_1$ to get $x_2=-7/24$. Now $(-7/24)^2+1=625/576$, voilà.

I might point out that when you’re seeking a $(q-1)$-th root of unity when the residue field is $\mathbb F_q$, the field with $q$ elements, there is a conceptually easier but computationally longer method, namely repetitive raising to the $q$-th power. Here, $i$ is a fourth root of unity, and $q=5$, so your sequence is $x_0=2$, $x_1=x_0^5=32$, $x_2=x_1^5$, etc., and you notice that $x_1^2+1\equiv0\pmod{5^2}$, $x_2^2+1\equiv0\pmod{5^3}$, etc. It’s easy enough to prove that this works.

Lubin
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