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For natural $a,b$, does $8a+5$ ever divide $b^2+8$ ?

It doesn't for $b$ up to $10^7$.

Couldn't find congruence obstructions for moduli up to $500$.

$b^2+8$ can be even.

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    $\begingroup$ Why I can't see who migrated this? $\endgroup$
    – joro
    Commented Jun 10, 2016 at 13:40

4 Answers 4

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No. $8a+5$ has a prime divisor $p$ congruent to 5 or 7 modulo 8 (since an odd natural number having only prime divisors congruent to 1 or 3 modulo 8 is itself congruent to 1 or 3). Next, -2 is quadratic non-residue modulo $p$, thus $(b/2)^2+2$ is not divisible by $p$.

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    $\begingroup$ $b$ can also be odd, so better make it "$-8$ is a quadratic nonresidue modulo $p$". $\endgroup$ Commented Jun 10, 2016 at 8:04
  • $\begingroup$ But $b$ is not odd modulo $p$:) $\endgroup$ Commented Jun 10, 2016 at 9:35
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    $\begingroup$ The counter-example $a = -6$ and $b=11$ in individ's answer below asks for this answer to be revised. $\endgroup$
    – Chris Wuthrich
    Commented Jun 10, 2016 at 11:50
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    $\begingroup$ @ChrisWuthrich $a$ is said to be natural $\endgroup$ Commented Jun 10, 2016 at 11:58
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    $\begingroup$ Yes, what I meant is that it would be nice to highlight that you have used this assumption. $\endgroup$
    – Chris Wuthrich
    Commented Jun 10, 2016 at 12:17
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Using the fact that $\mathbb{Z}[\sqrt{-2}]$ has unique factorisation, we can write:

$$ 8a + 5 \; \; | \; \; (b + 2 \sqrt{-2})(b - 2 \sqrt{-2}) $$

Now, the $\gcd$ of $b \pm 2 \sqrt{-2}$ divides $4\sqrt{-2}$. On the other hand, $8a + 5$ is not divisible by $\sqrt{-2}$, so any prime divisor of $8a + 5$ cannot divide both $b \pm 2 \sqrt{-2}$.

Consequently, $8a + 5$ has no real prime divisors, so we can write:

$$ 8a + 5 = b^2 + 2c^2 $$

Working modulo $8$, this is an immediate contradiction.

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  • $\begingroup$ You don't need positivity anywhere right? Negative counterexamples exist. $\endgroup$
    – joro
    Commented Jun 10, 2016 at 12:45
  • $\begingroup$ @joro, it is implicitly used: in negative case we can also have $8a+5=-(b^2+2c^2)$ $\endgroup$ Commented Jun 10, 2016 at 12:56
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For the equation.

$$b^2+8=(8a+5)c$$

You can write for example the following parameterization.

$$b=12p^2+5p+11$$

$$a=-(6p^2+p+6)$$

$$c=-(3p^2+2p+3)$$

$a,c - $ turn out negative.

Will make a replacement. We introduce the number. $k=\frac{t^2+8}{3}$

We will use the solutions of Pell's equation. $p^2-8s^2=1$

Knowing the first solution i $(p_0;s_0) - (3;1)$, you can find the rest on the previous formula.

$$p_2=3p_1+8s_1$$

$$s_2=p_1+3s_1$$

Now knowing this, you can write down the solutions themselves.

$$b=tp^2+(8k+3)ps+8ts^2$$

$$a=-(p^2+2tps+(8k-5)s^2)$$

$$c=-(kp^2+2tps+3s^2)$$

$t,p,s - $ can have any sign.

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  • $\begingroup$ Thanks. But very likely this is not complete parametrization or is it? $\endgroup$
    – joro
    Commented Jun 10, 2016 at 11:58
  • $\begingroup$ It is necessary to write other parameterization? $\endgroup$
    – individ
    Commented Jun 10, 2016 at 12:10
  • $\begingroup$ Yes, complete parametrization, otherwise these are special cases. $\endgroup$
    – joro
    Commented Jun 10, 2016 at 12:43
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You say you couldn't find a congruence obstruction for moduli up to $500$, but there is such an obstruction to having $-8 \equiv b^2 \bmod 8a+5$, revealed by Jacobi reciprocity. Since $-8$ is a unit modulo $8a+5$, so would be $b$, and therefore using Jacobi symbols we would get $(\frac{-8}{8a+5}) = (\frac{b^2}{8a+5}) = 1$. Positivity of $a$ is used in the meaning of the Jacobi symbol that was just written down and, more importantly, in computational rules for the Jacobi symbol that will now be used.

Since $8a+5\equiv 5 \bmod 8$ we have $(\frac{2}{8a+5}) = -1$, and $(\frac{-1}{8a+5}) = (-1)^{((8a+5)-1)/2} = 1$, so $(\frac{-8}{8a+5}) = (-1)^3 = -1$. This is a contradiction.

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    $\begingroup$ Thanks. You mean obstruction for positive $a$, since according to sage $11^2 \equiv -8 (\mod 8\cdot-6+5)$. $\endgroup$
    – joro
    Commented Jun 10, 2016 at 15:34
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    $\begingroup$ I had already pointed out in my answer where positivity is needed, since I noticed you keep bringing up this issue in comments to other people's answers. You don't have to clarify something that I already did from the start. $\endgroup$
    – KCd
    Commented Jun 10, 2016 at 15:43

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