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Could You give your ideas, your comment, or a referen for a conjecture as follows:

Consider $A, B, C$ be three positive integers numbers. By Fundamental theorem of arithmetic we write:

$A=a_1^{x_1}a_2^{x_2}...a_n^{x_n}$,

$B=b_1^{y_1}b_2^{y_2}...b_m^{y_m}$,

$C=c_1^{z_1}c_2^{z_2}...c_k^{z_k}$

Let $N=\min\{x_i, y_j, z_h \}$ where $1 \le i \le n, 1\ \le j \le m, 1\le h \le k$

For a positive integer $N_0 > 3$, there exist only finitely many triples $(A, B, C)$ of coprime positive integers, with $A+B=C$, such that: $N \ge N_0$

Edited: I repalced $N_0 \ge 3$ by $N_0 > 3$

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This conjecture is false, though only barely.

It should be true for $N_0 \geq 4$ by the ABC conjecture. But for $N_0 = 3$ there are infinitely many counterexamples such as $$ 191114642^3 + 4309182809^3 = 7^4 \, 321817873^3. $$ I obtained this by starting from the point $P_1 = (2:1:1)$ on the elliptic curve $x^3+y^3=7z^3$ (with origin $P_0 = (1:-1:0)$), and looking for a multiple $P_n = nP_1$ whose $z$-coordinate is divisible by $7$; such $n$ must exist because $7|z(P_n)$ iff $P_n$ reduces mod $7$ to $P_0$, and here it turned out that $n=7$ was the first example. Then any $n=7m$ works, so we get an infinite sequence of counterexamples.

The same construction works for any elliptic curve $x^3+y^3=Dz^3$ of positive rank.

P.S. I see that a simpler counterexample$$ 271^3 + 3^5 146^3 = 919^3 $$ was given by user459832 to the parallel question on Math Stackexchange. This corresponds to the third multiple of the generator $(2:1:1)$ on $x^3+y^3=9z^3$; here the first case of $3|z$ is small enough that we can still comfortably exhibit the next one, $$ 415280564497^3 + 676702467503^3 = 3^5 116223894220^3. $$

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  • $\begingroup$ Can You help me improving the conjecture? $\endgroup$ – Cố Gắng Lên Jul 4 '17 at 5:01
  • $\begingroup$ Dear Dr. @NoamD.Elkies if $N_0 \ge 4$ the conjecture can easily prove by abc conjecture? Or can using this conjecture to prove the abc conjecture? $\endgroup$ – Cố Gắng Lên Jul 4 '17 at 5:26
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    $\begingroup$ Your conjecture for $N_0 \geq 4$ is a special case of the ABC conjecture; that's what I was getting at in the second sentence of my answer. The ABC conjecture is much stronger, so I don't think that a proof of the $N_0 \geq 4$ part of your conjecture could be used to prove the full ABC conjecture. Still I don't think any techniques are known for solving even your more modest conjecture for any value of $N_0$, however large (except of course if Mochizuki's approach to the full ABC conjecture turns out to work). $\endgroup$ – Noam D. Elkies Jul 4 '17 at 5:31
  • $\begingroup$ I repalced $N_0 \ge 3$ by $N_0 >2$ $\endgroup$ – Cố Gắng Lên Jul 7 '17 at 8:14

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