If $A+B \mid A^5$ and $A+B \mid B^5$, what can be concluded?

Question

Assume $A+B \mid A^5$ and $A+B \mid B^5$, while all the variables are integers expect zero.

Can we prove that $A=B$?

This is my idea of the proof:

For every prime $p$ that $p\mid A+B$ there is $p \mid A$ and $p \mid B$.
Let $\gcd(A,B)=P$, so $A=Pw$ and $B=Pv$ such that $w$ and $v$ have no common factors with $P$. Then if $q\ne 1$, such that $q$ and $P$ are coprime, and $q\mid A+B$ then $q\mid w+v$, and hence $q$ divides both $A$ and $B$, and this is not possible.
Hence $A=P$ and $B=P$.

In this proof, I did not use the exponent $5$ at all. So I feel there is something wrong with the proof. Is there any thoerem that can be used here?

It’s not necessarily true; $A = 3, B = 6$ satisfies these conditions. — paulinho, Sep 21 '20 at 13:56
The two assumptions are redundant: $A + B \mid A^5$ is equivalent to $A + B \mid B^5$. — WhatsUp, Sep 21 '20 at 13:56
$(A+B) \mid (A^5+B^5)$ is always true. In fact $A^5+B^5 = (A+B)(A^4-A^3B+A^2B^2-AB^3+B^4)$. — player3236, Sep 21 '20 at 14:00
@DietrichBurde here the assumption is that "A+B divides A^5 and also divides B^5" so here this sentence should be correct. — Optima, Sep 21 '20 at 14:01
There's something that I felt - $B$ could be $0$ while $A$ could be something else (or vice versa). — Spectre, Sep 21 '20 at 14:05
@player3236 but here the assumption is $A+B | A^5$ and $A+B | B^5$ which is not always true. It is true only due to an assumption here. — Optima, Sep 21 '20 at 14:05
@Spectre true. Zeros should be excluded. I will edit the question. — Optima, Sep 21 '20 at 14:06
Good then ! I felt we required Mr Zero to come into the question..... — Spectre, Sep 21 '20 at 14:06
The aforementioned counterexample holds since $3 \mid 1+2$ ($p \mid w+v$). This possibility was not (and cannot be) accounted for in your proof. — player3236, Sep 21 '20 at 14:09
If $A +B | A^5$ or $A +B | B^5$, can't $A +B $ be a prime factor of either of the two ? — Spectre, Sep 21 '20 at 14:09
Or say, $A^5 $ or $B^5$ can be factorised and any of the factors of those could be cancelled out (I mean, divided off) using $A + B$, right? — Spectre, Sep 21 '20 at 14:17
@Spectre but is not it a case of v+w in the second bullet point? The idea is that if A and B have a factor other than A + B, then that factor is common between them. So what ever factors A has, B has the exact same factors. Thus A and B are the same. Makes sense do you think? — Optima, Sep 21 '20 at 14:24
Well, I am 15 and I have not much sense to talk the kind of talks you people are having now, so I may be a bit incapable in understanding the whole topic. — Spectre, Sep 21 '20 at 14:26
@Optima, I have also added a few other opinions of mine in the comments below my answer. — Spectre, Sep 21 '20 at 15:13

Bill Dubuque · Accepted Answer · 2020-09-21T19:59:06.093

3

As mentioned there are easy counterexamples. We can characterize the solutions as follows. By the gcd Universal Property & Freshman's Dream we have

$\begin{align} A+B\mid A^5,B^5\!\iff\! A+B&\mid (A^5,B^5)=(A,B)^5\\[.2em] \iff a\ +\ b&\mid\, c^4\ \ \text{by dividing prior by }\,c := (A,B), \ {\rm with }\,\ a,b = \frac{A}c,\frac{B}c \end{align}$

So let $\,c\,$ and coprime $\,a,b\,$ be solutions of $\,a+b = c^4.\,$ Then $\,A\!+\!B\mid A^5,B^5$ for $\,A,B = ac,bc$.

e.g. let's view the first counterexample in the comments this way

$$\begin{align} 1+2&\mid 3^4\\[.2em] \iff\ 3+6&\mid 3^5 = (3,6)^5 = (3^6,6^5)\\[.2em] \iff\ 3+6&\mid 3^5,6^5\\[.3em] \text{for a bigger example }\ \ \ 40\ +\ 41\ &\mid\ 3^4\\[.2em] \iff\ 120+123&\mid 120^5, 123^5 \end{align}\qquad\qquad\qquad\qquad\quad $$

edited Sep 21 '20 at 19:59

answered Sep 21 '20 at 17:14

Bill Dubuque

272,048

You are better than me.... I bow down, sire ! :) – Spectre Sep 21 '20 at 17:36
What about my answer ? – Spectre Sep 21 '20 at 17:36
Is there anything worth understanding from it ? – Spectre Sep 21 '20 at 17:37
@Spectre I haven't had a chance to read it, but you may find it interesting to view your examples from the above viewpoint. – Bill Dubuque Sep 21 '20 at 17:43
I'll try it... thanks for the advice... – Spectre Sep 21 '20 at 17:44
But was my answer sensible, @BillDubuque ? – Spectre Sep 21 '20 at 17:45

Spectre · Answer 2 · 2020-09-21T17:57:19.103

1

NB : I'll be using $(x,y)$ to denote the values of $A$ and $B$ as $(A,B)$.

"Suppose $A$ and $B$ are integers other than $0$, such that $A + B|A^5$ and $A + B|B^5$, what inferences can be drawn from this ?" - this is what has been asked.

If the above conditions have to be satisfied, $A^5$ and $B^5$ should have $A + B$ as a factor (or some other number divisible by that sum) . This means that $A = 0$ or $B = 0$ , which is invalid as per the question. Therefore, we can consider values like $(3, 6)$ (as @paulinho suggested) , but at the same time, we have exceptions : $(2,9) [\because 11 \nmid2^5 , 11 \nmid 9^5]$ and such values. So, maybe, $A +B$ should be some power of primes to satisfy the condition (as in the example $(3,6) : 3 + 6 = 9 = 3^2$; or even $(2,6):2 + 6 = 8 = 2^3$).

On the whole, if $p$ is a prime, a pair of integers $(p, pn)$ [for some $n$ such that $p \mid (n + 1)$ and $(n + 1) = p^m$ for some $m < 5$] satisfies the conditions.

PS : For a better answer, look for the one written by @BillDubuque. I think that his answer is more sensible and more understandable than mine.

edited Sep 21 '20 at 17:57

answered Sep 21 '20 at 14:45

Spectre

1,573

I believe that I am clear enough to you ... – Spectre Sep 21 '20 at 14:46
Note : the above condition that you have to take a prime needn't be followed. I wrote what I had ended up in. You may find counterexamples and add that to the answer to make it more OK .... – Spectre Sep 21 '20 at 15:06
@BillDubuque, I hope you'll edit your comment.. a few typos came up, I guess.. – Spectre Sep 21 '20 at 17:37
Never mind... I have put my comments beneath your answer, @BillDubuque ... – Spectre Sep 21 '20 at 17:43

If $A+B \mid A^5$ and $A+B \mid B^5$, what can be concluded?

2 Answers2