Smallest prime that divides $n^2+5n+23$

Question

Find the smallest prime that divides $n^2+5n+23$ for some integer $n$. I thought taking all primes less than 29 one by one and then solving the equations and then some manipulation. (Like $n^2+5n+23=2k$). But it didn't help. What else can I do?

Answer 1

If you are unfamiliar with the theory of quadratic residues, then you need to do a bit more work. To exclude the possibility of a prime $p$ from occurring as a factor, you need to test it with all the cases $n=0,1,2,\ldots,p-1$. Do you see why this suffices?

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It is easier with a bit of theory of quadratic residues. I use the properties of the Legendre symbol $\left(\frac a p\right)$ that is $+1$ if $a$ is a QR modulo $p$ and $-1$ if not.

We can use the "observation" that, given a prime $p$, there exists an integer $n$ such that $p\mid n^2+5n+23$ if and only if the equation $n^2+5n+23=0$ has a solution in the field $\Bbb{Z}_p$ if and only if the discriminant $$ D=5^2-4\cdot23=-67 $$ is a quadratic residue modulo $p$. This works whenever $p$ is odd, and the case $p=2$ is easy to handle.

Split into two cases.

I) $p$ congruent to $1$ modulo $4$: Here we know that $\left(\frac{-1}p\right)=1$, so we want to know, if $\left(\frac{67}p\right)=1$. $$ \begin{aligned} \left(\frac{67}5\right)&=\left(\frac{2}5\right)=-1,\\ \left(\frac{67}{13}\right)&=\left(\frac{2}{13}\right)=-1,\\ \left(\frac{67}{17}\right)&=\left(\frac{67-3\cdot17}{17}\right)=\left(\frac{4^2}{17}\right)=1,\\ \end{aligned} $$ so $p=17$ is the smallest possible factor. When evaluating $\left(\frac 2p\right)$ use either brute force testing or the appropriate extension of the law of quadratic reciprocity.

II) $p$ congruent to $3$ modulo $4$: Here we know that $\left(\frac{-1}p\right)=-1$, so we want to know, if $\left(\frac{67}p\right)=-1$.

$$ \begin{aligned} \left(\frac{67}3\right)&=\left(\frac{1}3\right)=1,\\ \left(\frac{67}{7}\right)&=\left(\frac{4}{4}\right)=1,\\ \left(\frac{67}{11}\right)&=\left(\frac{67-6\cdot11}{11}\right)=\left(\frac{1}{11}\right)=1,\\ \end{aligned} $$ so $p=3,7,11$ cannot occur as prime factors. The next prime in this residue class is $p=19$, which is larger than the smallest prime factor $17$ from case I.

So seventeen it is.

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Answer 2

Let's have a look at this systematically. It is trivial that $n^2+5n+23$ is odd, so the prime $2$ is out of the picture.

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A general observation about quadratic equations modulo $p$

If $n^2+an+b\equiv 0$ modulo some prime $p$ - say for the integer $r$ then $$(n-r)(n+a+r)=n^2+an-r^2-ar\equiv n^2+an+b$$ (because $b\equiv -r^2-ar$) so that the equation has a factorisation modulo $p$.

So it makes sense to talk of a root $r$ mod $p$ and this is always associated with a factorisation of the quadratic. (More general cases can be shown)

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Now, since $2$ is out of the picture, multiply by $4$ to work with $4n^2+20n+92=(2n+5)^2+67$, which has a root modulo an odd prime $p$ if and only if $-67$ is a square mod $p$.

Then we test primes. We have $-67 \equiv 2 \bmod 3; 3 \bmod 5; 3\bmod 7; 10\bmod 11 \dots$ or use quadratic reciprocity.

Answer 3

This is not a complete answer, but instead provides some evidence that there won't be a simple solution to this problem. Running FactorInteger[n^2 + 5*n +23] up to $n = 25$ in Mathematica yields the following results. $$\begin{array}{c|l} n & n^2 + 5n +23 \\\hline 1 & 29 = 29 \\ 2 & 37 = 37 \\ 3 & 47 = 47 \\ 4 & 59 = 59 \\ 5 & 73 = 73 \\ 6 & 89 = 89 \\ 7 & 107 = 107 \\ 8 & 127 = 127 \\ 9 & 149 = 149 \\ 10 & 173 = 173 \\ 11 & 199 = 199 \\ 12 & 227 = 227 \\ 13 & 257 = 257 \\ 14 & 289 = 17^2 \\ 15 & 323 = 17\times19 \\ 16 & 359 = 359 \\ 17 & 397 = 397 \\ 18 & 437 = 19\times23 \\ 19 & 479 = 479 \\ 20 & 523 = 523 \\ 21 & 569 = 569 \\ 22 & 617 = 617 \\ 23 & 667 = 23\times29 \\ 24 & 719 = 719 \\ 25 & 773 = 773 \\ \end{array}$$

Answer 4

Hint: You can change 23 to 19 + 4 so it will become $(n+1)(n+4)+19$

btw, the answer is not 19