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Let $x$ be an odd natural number. Show that $x^2+2 \equiv 3 \mod 4$ and deduce that there exists a prime $p$ with $p|x^2+2$ and $p \equiv 3 \mod 4$.

For the first part I would assume $$x \equiv 1 \mod 2$$ $$x^2 \equiv 1^2 \mod 2^2$$ $$x^2 +2 \equiv 1^2+2 \mod 2^2$$ $$x^2 +2 \equiv 3 \mod 4$$

Can someone briefly tell me if this is correct and if it is a rule that squaring the number on the left means you square both numbers on the right?

Also I don't know how to start with the second part.

Note: This question is to do with rings in general.

user499701
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    This is not correct. For example, $6\equiv1\mod{5}$ but $36\not\equiv1\mod{25}$ – Peter Foreman Jul 01 '19 at 17:37
  • @PeterForeman I thought it looked like a funny rule. Do you know how to show this otherwise? – user499701 Jul 01 '19 at 17:39
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    If $2|x-1$ then $2|x+1$ so $4|(x+1)(x-1)=x^2-1=x^2+2-3$ – J. W. Tanner Jul 01 '19 at 17:41
  • @Peter But that's not the correct generalization to primes $,p>2.,$ Instead, see my answer. – Bill Dubuque Jul 01 '19 at 17:56
  • You should be able to prove that odd squares are $\equiv 1 \bmod 8$, which is stronger than you need. Since $x^2+2$ is odd what happens if none of the factors have the form given? – Mark Bennet Jul 01 '19 at 18:04
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    @BillDubuque I was just answering the part about "a rule that squaring the number on the left means you square both numbers on the right" although it may be true that $\text{mod }2$ this works, it is not true in general. – Peter Foreman Jul 01 '19 at 18:12

4 Answers4

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Assuming $x$ is odd, $x\equiv1\pmod2$ so $2|x-1$ so $2|(x-1)+2=x+1,$

so $4|(x-1)(x+1)=x^2-1=x^2+2-3, $ so $ x^2+2\equiv3\pmod4$.

Let $p$ be a factor of $x^2+2$. $p$ must be odd because $x$ and therefore $x^2+2$ is.

If all such factors were $\equiv1\pmod4$ then their product would be $\equiv1\pmod4$, a contradiction.

So $x^2+2$ has a prime factor $\equiv3\pmod4$.

J. W. Tanner
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It is the special case $\,n = 2\,$ below.

Theorem $\ \ \color{#0a0}{a\equiv b}\pmod{\!n}\,\Rightarrow\, a^{\large n}\equiv b^{\large n}\pmod{\!n^{\large 2}}\ $ for all integers$\ a,b,n\,$ with $\,n\ge 0$.

Proof $\ \ \ a^{\large n}-b^{\large n} = (a\!-\!b)f(a,b)\,$ for $\, f(a,b) = a^{\large n-1}\!+ a^{\large n-2}b\ +\ldots +a b^{\large n-2}\!+b^{\large n-1}$

$\!\bmod\color{#c00}n\!:\,\ \color{#0a0}{a\equiv b}\,\Rightarrow\, f(\color{#0a0}a,b)\equiv f(\color{#0a0}b,b)\equiv \color{#c00}n\,b^{\large n-1}\equiv 0\, $ by the Polynomial Congruence Rule.

Hence $\,n\mid a\!-\!b, f(a,b)\,\Rightarrow\, n^{\large 2}\mid (a\!-\!b)f(a,b) = a^{\large n}-b^{\large n}$

Remark $ $ Alternatively we can prove it via the Binomial Theorem or Double Root Test, e.g. see here. See also LTE = Lifting The Exponent

Bill Dubuque
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For a more quick way: Notice, $x^2+2$ is always odd, as $x$ is odd. Therefore, all prime divisors of $x^2+2$ are either of form $4k+1$ or $4k+3$. Now, if all of them are of form $4k+1$, then simply check that $x^2+2\equiv 1\pmod{4}$. However, since $x$ is odd, $x^2+2\equiv 3\pmod{4}$, which is yielding a contradiction.

TBTD
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First, you can square both sides in the congruence relation but you must not square the number that you are "moding".

If $x$ is odd, then $x\equiv1$ or $3$ (mod $4$). So, by some calculations,

$$x^2\equiv1(\mod 4)$$ $$x^2+2=3(\mod 4)$$

Then, for the next part, we will use prove by contradiction.

First, we can see that all of the factors of $x^2+2$ is odd because $x^2+2$ is odd.

If $x^2+2$ has no prime factor which has a remainder of $3$ when divided by $4$, then $x^2+2$ should have only prime factors which has a remainder of $1$ when divided by $4$. As the product of any number of numbers that leaves a remainder of $1$ when divided by $4$ also has a remainder of $1$ when divided by $4$, so it contradicts that $x^2+2\equiv3 (\mod 4)$

$\therefore x^2+2 $ has at least one factor that leaves a remainder of $3$ when divided by $4$.

Culver Kwan
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