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Among $168$ prime numbers in range $1$ to $10^3$, there are $84$ prime numbers $n$ such that: $p^k> n.rad(p^{k+1}−n)$ where $1 \le n<p$ and $k=2,3,4$. There are also $84$ prime numbers $n$ such that: $p^k> (n.rad(p^{k+1}−n))^{1.00001}$ where $1 \le n<p$ and $k=2,3,4$. The ratio $\frac{84}{168}=0.5$ is high.

Question 1: Are there infinitely many primes $p$, positive integers $k, n$ such that $1 \le n < p$ and $p^k > n.rad(p^{k+1}−n)$?

Question 2: If the answer to question 1 is yes, is there $\varepsilon > 0$ such that there are infinitely many primes $p$, positive integers $k, n$ such that $1 \le n < p$ and $p^k > (n.rad(p^{k+1}−n))^{1+\varepsilon}$?

Question 3: If the answer of question 2 is yes, then is the ABC conjecture false? if we let $A=p^{k+1}-n$, $B=n$

$A+B=p^{k+1}-n+n=p^{k+1}=p.p^k>p.(n.rad(p^{k+1}-n))^{1+\varepsilon} \ge (p.n.rad(p^{k+1}-n))^{1+\varepsilon_0}=rad(ABC)^{1+\epsilon_0}$

List of $84$ pair $(p, n)$ as follows $p=A(2i-1)$, $n=A(2i)$ for $i=1,2,...,84$

Examples:

$13^2-10.rad(13^3-10)=139>0$

$23^2-17.rad(23^3-17)=19>0$

$37^2-28.rad(37^3-28)=949>0$

$A=$ [13 10 23 17 37 28 73 55 107 43 137 89 181 136 191 89 281 41 313 1 337 253 353 89 379 3 383 287 433 17 433 325 467 5 541 406 563 422 631 31 769 577 811 83 863 647 937 703 3 1 5 1 7 1 17 1 19 1 31 1 41 1 53 1 73 46 97 1 107 1 127 1 127 19 131 121 163 1 181 49 193 1 199 1 239 1 241 1 251 1 257 1 271 1 307 1 331 49 337 1 443 1 449 1 487 1 557 1 577 1 593 1 599 193 751 1 797 49 821 561 881 385 907 817 937 289 983 289 3 1 23 2 29 5 43 15 47 7 103 7 163 82 197 17 211 3 229 24 251 1 277 8 281 249 283 43 449 193 487 1 563 11 821 5 827 11 853 5]

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  • $\begingroup$ If you need it, one can to write the central dot of a product $a\cdot b$ if one type in the corresponding formula \cdot , also is required a blank space between the string cdot and the second operand if the second operand is a string of letters \cdot rad $\endgroup$ – user142929 Sep 11 at 9:30
  • $\begingroup$ Maybe, there are more and more inequalities in question 1 with the 168 primes. Because for fast test, In my calculation when I take first $n$ such that the inequality. I omit the number bigger $n$. $\endgroup$ – Đào Thanh Oai Sep 11 at 15:21
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    $\begingroup$ @user142929 I use matlab to test the idea $\endgroup$ – Đào Thanh Oai Sep 11 at 15:27
  • $\begingroup$ You're allowing $k=1$? Surely there are lots of examples where $p>2n$ and $p^2-n$ is a power of $2$. $\endgroup$ – Greg Martin Sep 11 at 16:49
  • $\begingroup$ Dear @GregMartin $k \ge 2$ and $k+1 \ge 3$, In the caculatation, I take only $k=2, 3, 4, 5$ $\endgroup$ – Đào Thanh Oai Sep 12 at 1:16

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