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Show that for all integers $j$, if $d$ is an integer such that $j \mid k+8$ and $d \mid k^2+1$ then $j \mid 65$.

I'm having a lot of trouble, with solving this proof. I first rewrote the equations as,

$$\begin{align*} aj &= k + 8, \\ bj &= k^2 + 1. \end{align*} $$

Then by adding them together,

$$\begin{align*} aj + bj &= k^2 + k + 9, \\ (a+b)j &= k^2 + k + 9. \end{align*}$$

But I am not sure how to proceed from here. Any help would be greatly appreciated.

Lenehan
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2 Answers2

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From $d \mid n + 8$, you have

$$\begin{equation}\begin{aligned} & n + 8 \equiv 0 \pmod d \\ & n \equiv -8 \pmod d \\ & n^2 \equiv 64 \pmod d \\ & n^2 + 1 \equiv 65 \pmod d \end{aligned}\end{equation}\tag{1}\label{eq1A}$$

Since $d \mid n^2 + 1$, from \eqref{eq1A}, you get $n^2 + 1 \equiv 65 \equiv 0 \pmod d \implies d \mid 65$.

John Omielan
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$d \mid (n+8) \Rightarrow d \mid (n+8)^2$

So: $d \mid (n^2 + 16n + 64) \tag{1}$

Also: $d \mid (n^2 + 1) \tag{2}$

So $d$ divides this difference:

$d \mid (n^2 + 16n + 64) - (n^2 + 1) \Rightarrow d\mid (16n + 63) \tag{3}$

Now since $d \mid (n+8)$ we get:

$d \mid 16(n+8) \tag{4}$

$d \mid (16n+128) \tag{5}$

Finally from $(5)$ and $(3)$ we get that $d$ divides this difference:

$(16n+128) - (16n + 63) = 65$

peter.petrov
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  • This proof is correct, it shouldn't be downvoted. – peter.petrov Apr 17 '20 at 00:03
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    Usually, to denote that "$a$" divides "$d$", you write $a \mid d$ not $a / d$. Note the use of the LaTeX command "\mid" (or "|" if you're lazy and don't care about spacing). I'm not the person who downvoted your proof. – eeen Apr 17 '20 at 00:06
  • Ahh, thank you! I understand how to approach to the question now – Lenehan Apr 17 '20 at 00:08
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    @Lenehan Yeah, the approach is first get rid of the $n^2$ term, then of the $n$ term which means you will be left with just a number (65 in this case). – peter.petrov Apr 17 '20 at 00:09
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    Denote $t = gcd(n+3, n^2+1)$. Take $(n+3)$ to power $2$. Then $t$ divides $(n+3)^2 - (n^2+1) = 5n+5$. So $t | (5(n+1))$ Can you go on from here? $5$ does not divide $(n+3)$ because it divides $n$. The same argument applies to the number $2$. So $5$ and $n+3$ are coprime and so $5$ and $t$ are coprime. So $t | (n+1)$. But also $t | (n+3)$. So... $t | 2$ Now again obviously $t$ cannot be $2$ so $t$ must be $1$. – peter.petrov Apr 17 '20 at 00:30
  • Yes, thank you! – Lenehan Apr 17 '20 at 00:34
  • @peter.petrov Note you stated in your comment above that $(n+3)^2 - (n^2 + 1) = 5n + 5$, but actually it's $6n + 8$ on the right side instead. – John Omielan Apr 17 '20 at 00:39
  • @JohnOmielan I really liked the approach of using congruences and modular arithmetic to this question, just wondering if a similar methodology can be applied to part 2? – Lenehan Apr 17 '20 at 01:07
  • @Lenehan The concept of congruences and modular arithmetic can basically always be applied to questions like yours, although it's not always necessarily the easiest or best method to use. However, in your case, let $d = \gcd(n + 3, n^2 + 1)$. Then $n + 3 \equiv 0 \pmod d \implies n \equiv -3 \pmod d \implies n^2 + 1 \equiv 9 + 1 \pmod d$. This means $d \mid 10$. However, since $10 \mid n$, then $10 \not\mid n + 3 \implies d \not\mid n + 3$ &, thus, $d = 1$, i.e., $n + 3$ & $n^2 + 1$ are coprime. Also, you can use the Polynomial congruence rule as member "gone" suggested in a question comment. – John Omielan Apr 17 '20 at 01:14
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    @Lenehan You're welcome for the explanation. However, I didn't explain the end part properly. I should have stated something like that since $d \mid 10$ and $10 \mid n$, then $d \mid n$. Since $\gcd(3,10) = 1$, then if $d \gt 2$, you have $d \not\mid n + 3$. However, since $d = \gcd(n + 3, n^2 + 1)$, then $d \mid n + 3$, so the only possibility is $d = 1$, i.e., $n + 3$ and $n^2 + 1$ are coprime. I hope this is more clear. – John Omielan Apr 17 '20 at 01:30
  • Simpler: $,d,\mid, n^2!+!1-(n!-!8)(n!+!8) = 1!+!8^2.,$ More conceptually $\bmod d!:\ i:= n,$ is a sqrt of $,-1$ so $,0\equiv \omega := i-8,\Rightarrow,0\equiv \omega,\bar\omega = 65\ $ @Lenehan – Bill Dubuque Apr 17 '20 at 03:55