Is there a way to show that $5a(a+1)\over 3a+4$ $\notin \mathbb{N}$ when $a\in\mathbb N\setminus\{2\}$?

Question

Is there a way to show that $5a(a+1)\over 3a+4$ $\notin \mathbb{N}$ for $a\in \mathbb{N}$ (except when $a=2$)?

The expression has a few different forms, but I don't see how to show this. Any hints are greatly appreciated.

Follows immediately via (scaled) Euclidean algorithm in the dupes, cf. this comment below. $\ \ $ — Bill Dubuque, May 27 '24 at 22:02
Unlike this question, the proposed duplicates ask for gcd. For this problem, even once the gcd is calculated, that is not the answer to the question -- there is more work to do -- and there are alternative answers (see below) not involving calculating the gcd — J. W. Tanner, May 28 '24 at 02:09

score 2 · Accepted Answer · answered May 26 '24 at 22:38

2

Yes, using polynomial division (multiplying by $9$ to maintain coefficients integers),

we see that $45a(a+1)=(3a+4)(3a-1)5+20,$

so if $3a+4$ divides $5a(a+1)$, then $3a+4$ divides $20$.

Can you take it from here?

answered May 26 '24 at 22:38

J. W. Tanner

This makes sense. Thank you. – ddswsd May 26 '24 at 22:48
@ddswsd special case of scaled Euclidean alg. in dupe, i.e. scaling 2nd gcd arg by $,c = \color{#0a0}{3\cdot 3},$ we get $$(n,5a(a!+!1)) = (n,5(\color{#0a0}{3a})(\color{#0a0}{3a}!+!3)) = (n,5(\color{#c00}{-4})(\color{#c00}{-4}!+!3))=(n,20)\qquad$$
since $\bmod n!=!\color{#0a0}{3a}!+!\color{#c00}{4}!:\ \color{#0a0}{3a}\equiv \color{#c00}{-4}.\ $ [note $,(n,c)=1,$ by $,(n,3)=(4,3)=1$] $\ \ $
– Bill Dubuque May 27 '24 at 21:59
Please strive not to post more (dupe) answers to dupes of FAQs. This is enforced site policy, see here. – Bill Dubuque May 27 '24 at 22:01

score 2 · Answer 2 · edited May 26 '24 at 23:48

2

$3a+4=3(a+1)+1$ and $a+1$ are prime to each other.

So if $\frac{5a(a+1)}{3a+4} \in \mathbb N$, then $2 > \frac{5a}{3a+4} \in \mathbb N$

Then the only option for $\frac{5a(a+1)}{3a+4} \in \mathbb N$ is that $\frac{5a}{3a+4} = 1$, which is $a=2$

edited May 26 '24 at 23:48

J. W. Tanner

answered May 26 '24 at 22:47

MoonKnight

Please strive not to post more (dupe) answers to dupes of FAQs. This is enforced site policy, see here. – Bill Dubuque May 27 '24 at 22:01

2 Answers2