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This conjecture is obviously inspired by the abc-conjecture:

Let $\gcd(a,b)=1$ then $\operatorname{rad}((a+b)ab(ab+a+b))> ab+a+b$

I am not asking for a proof, just for possible counterexamples, if they exist. I checked this with the computer for some numbers, and didn't find any counterexample.

What I checked so far $(\gcd(a,b)=1)$:

  1. $1 \le a,b \le 1000$
  2. $a=1$, $1 \le b \le 10^6$
  3. $1 \le m \le 10^6$, $a=m,b=m+1$

Heuristic that this is true for infinetly many $b$: If $p\neq 2$ is a prime, then set $b = \frac{p-1}{2}, a = 1$. Then $\operatorname{rad}((a+b)ab(ab+a+b)) = \operatorname{rad}(\frac{p-1}{2}\frac{p+1}{2}p) > p = ab+a+b$

Another way to prove that there are infinitely many $(a,b)$ which fulfill the conjecture: Choose some $a \in \mathbb{N}$. Since $\gcd(a,a+1)=1$, by the (https://en.wikipedia.org/wiki/Dirichlet%27s_theorem_on_arithmetic_progressions) Dirichlet theorem on arithmetic progression there are infinitely many primes of the form $p = b(a+1)+a = ab + a +b$. Then necessarily $\gcd(a,b)=1$, otherwise if $g=\gcd(a,b)$ $a = g a_1$ and $b=g b_1$ then $g | p$ and hence $g=p$, which is impossible since $p = g ( a_1 b_1+a_1+b_1)$ and we must have $3 \le a_1b_1+a_1+b_1=1$, which can not work. Then $\operatorname{rad}((a+b)ab(ab+a+b)) = \operatorname{rad}((a+b)ab \cdot p) > p = ab + a +b$

Edit: If someone finds another way to produce infinitely many tuples $(a,b)$ which fulfill the conjecture, that would also be interesting.

Second Edit: Related question: https://mathoverflow.net/questions/343245/other-examples-of-irreducible-similarities-over-the-natural-numbers

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  • $\begingroup$ I add here the Wikipedia article defining the radical of an integer $\operatorname{rad}(n)$. Good week. $\endgroup$
    – user243301
    Aug 5, 2017 at 9:18
  • $\begingroup$ It's just a thougt , you have $x+y=z$ and $rad(x y z ) > z $ such that $x,y$ are coprimrs, in a sense for $rad$ to be smaller, you need a lot of powers, so it seems in this way that you will not find counter-example (may be few). $\endgroup$
    – Ahmad
    Aug 5, 2017 at 9:26
  • $\begingroup$ There must be a typo or something, because there are a lot of counterexamples. For 0<a,b<1000 there are $60887$ counterexamples, for example $(41,97),(42,95),(42,97),\dots$. $\endgroup$
    – Lehs
    Aug 5, 2017 at 10:03
  • $\begingroup$ @Lehs: Thank you for your comment: How is $(41,97)$ a counterexample? $a=41,b=97$ I get $\operatorname{rad}((a+b)ab(ab+a+b)) = 2 \cdot 3 \cdot 5 \cdot 23 \cdot 41 \cdot 97 \cdot 823$ and $ab+a+b = 5 \cdot 823$. What is your computation? $\endgroup$
    – user276611
    Aug 5, 2017 at 10:12
  • $\begingroup$ @Lehs: I don't think this is correct. Even if it was, this number is still greater then $ab+a+b=5 \cdot 823$. Since $rad(ab+a+b)=ab+a+b=5\cdot 823$ the factor $5\cdot 823$ must divide $rad((a+b)ab(ab+a+b))$ $\endgroup$
    – user276611
    Aug 5, 2017 at 10:29

4 Answers 4

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This is not an answer, but a community wiki to avoid a lot of comments.

What I have tested without failure, so far:

$a=2$ and $b=1,\dots, 1,000,000$
$0<a,b<10,000$

Also tested without failure:

  1. $3 \le p \le 10^6 $, $p$ prime. $a = 1$, $b = \frac{p^2-1}{2}$
  2. $ p \neq q, p,q \le 10^4$, $p,q$ primes, $a=p$, $b=q$
  3. $ 3 \le p \le 10^6$, $p$ prime, $a=\frac{p-1}{2},b=\frac{p+1}{2}$
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One easy way to get infinitely many true cases is to assume $a$ and $b$ are squarefree. Then the radical of $(a+b)ab(ab+a+b)$ is necessarily a multiple of $ab$. Both $a+b$ and $ab+a+b$, being coprime to $ab$ and mutually coprime, must contribute some nontrivial prime factor to the radical. Therefore the radical is at least $2 \cdot 3 \cdot ab$, which is strictly greater than $ab+a+b$.

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  • $\begingroup$ Thanks for your answer: I have a question: The $2$ comes from a nontrivial factor of $a+b$ and the $3$ comes from a nontrivial factor of $ab+a+b$? $\endgroup$
    – user276611
    Aug 6, 2017 at 8:32
  • $\begingroup$ @stackExchangeUser $a+b$ contributes at least one prime factor, and $ab+a+b$ contributes at least one different prime factor. The product of these two prime factors is at least 6, but we don't have control over a precise correspondence. $\endgroup$ Aug 6, 2017 at 8:36
  • $\begingroup$ Ok thanks, that makes sense. Thank you again for your answer! $\endgroup$
    – user276611
    Aug 6, 2017 at 8:37
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Exhaustive list of triples $(A,B,C)$ (for $C<10^{18}$) such that $$A+B=C,\\rad(ABC)<C,\tag{1}$$ is accessible by the link Bart de Smit/ABC triples/by size, file abctriples_below_1018.gz .

If such pair-counterexample $(a,b)$ exists (as described in the question), then $GCD(a+b,ab)=1$, and if construct $$\begin{array}{l}A=a+b,\\B=ab, \\C=ab+a+b;\end{array}\tag{2}$$ then $(1)$ must be true for constructed $(A,B,C)$.

But there are no triples $(A,B,C)$ from the database above of the form $(2)$.

(If numbers $A,B$ can be written in the form $A=a+b,B=ab$, then polynomial $f(x)=x^2-Bx+A$ has $2$ positive integer roots.)

So, to find counterexample $(a,b)$, one needs search in the range $ab+a+b\ge 10^{18}$. $\color{#E0E0E0}{Hopeless...}$

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  • $\begingroup$ @Olge567: Did you check that there are no such triples in the database? $\endgroup$
    – user276611
    Aug 7, 2017 at 8:21
  • 1
    $\begingroup$ @stackExchangeUser: yes, and didn't find any. $\endgroup$
    – Oleg567
    Aug 7, 2017 at 8:22
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This question is related to the following conjecture: $$(x+y)^2<\operatorname{rad}\bigl[xy(x+y)(x^2+xy+y^2)\bigr]$$

Assuming $a<b$, we can consider other triples: $$(a,b(a+1),a+b+ab) \tag{1}$$ $$(b,a(b+1),a+b+ab) \tag{2}$$ $$(2(a+b),ab-a-b,a+b+ab) \tag{3}$$

(1) and (2) can be an ABC-triple, although for (2) I only found two examples checking the mentioned ABC-database: $(2^{16}.5^2,3^6.41.89^2.449,571^4)$ and $(2^4.5^6,53^2.89.499^2,3^2.7^2.109^4)$.

(3) seems to have the same property as the above conjecture: $$ab+a+b < \text{rad} \big( 2(a+b)(ab-a-b)(ab+a+b) \big)=\text{rad} \big( 2(a+b)(a^2b^2-(a+b)^2)\big)$$

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