Prove that for all prime numbers $n$ larger than 3 that $(9n+1)^2-(n+9)^2$ is divisible by 1920

Question

Prove that for all prime numbers $n$ larger than $3$ that $(9n+1)^2-(n+9)^2$ is divisible by $1920$.

Hi there, I've tried this problem for the past two days on and off but I'm getting increasingly frustrated as I think it's a really simple problem with a simple solution, but I'm having trouble connecting the dots. I've tried looking at congruencies and sub-cases to no avail, I've researched similiar problems and they usually look into proving divisibility with $128$ and $15$ ($3,5$). Any help's appreacited, thanks.

Hint: Every prime larger than $3$ is of the form $n = 6k \pm 1$. — Martin R, Mar 02 '22 at 20:24
Does this answer your question? For any prime $p> 3$, why is $p^2-1$ always divisible by 24?. $(9n+1)^2-(n+9)^2=80(n^2-1)$ and $24\cdot 80=1920$. — Dietrich Burde, Mar 02 '22 at 20:32

score 0 · Answer 1 · answered Mar 02 '22 at 20:32

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Any prime number $n\ge 5$ is of the form $6k-1$ or $6k+1$, hence $24\mid n^2-1\iff n^2-1=24k$ For any prime $n$. Now, $$(9n+1)^2-(n+9)^2=(8n-8)(10n+10)=80(n-1)(n+1)=80(n^2-1)=1920k$$

answered Mar 02 '22 at 20:32

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score 0 · Answer 2 · answered Mar 02 '22 at 20:32

Use the formula $a^2-b^2 = (a-b)(a+b)$ to get $$N = (9n+1)^2 - (n+9)^2 = (8n-8)(10n+10) = 80 \cdot (n-1)(n+1) = 80 \cdot (n^2-1)$$ $\textbf{1.}$ Clearly $5 \mid N$

$\textbf{2.}$ Since $3 \nmid n$, it follows that $n^2 \equiv 1 \pmod 3$, so $3 \mid N$

$\textbf{3.}$ Since $n$ is odd $n^2 \equiv 1 \pmod 8$, so $128 \mid N$

Prove that for all prime numbers $n$ larger than 3 that $(9n+1)^2-(n+9)^2$ is divisible by 1920

2 Answers2