Skip to main content
MathOverflow

Return to Answer

links, formatting
Source Link
edit approved Jul 5, 2020 at 2:35
Gerry Myerson
edit approved Jul 5, 2020 at 2:35
Gerry Myerson
  • 1.1k
  • 16
  • 16

Felipe refers to my first ever paper (mathscinet.ams.org/mathscinet-getitem?mr=789713mathscinet.ams.org/mathscinet-getitem?mr=789713) from 1985 ! However I have a more recent paper that gives a better result along the lines asked for (mathscinet.ams.org/mathscinet-getitem?mr=2997580mathscinet.ams.org/mathscinet-getitem?mr=2997580) which shows that every 2^n-1$2^n-1$, with n<>1$n\ne1$ or 6$6$, has a primitive prime factor that divides it to an odd power.

Felipe refers to my first ever paper (mathscinet.ams.org/mathscinet-getitem?mr=789713) from 1985 ! However I have a more recent paper that gives a better result along the lines asked for (mathscinet.ams.org/mathscinet-getitem?mr=2997580) which shows that every 2^n-1, with n<>1 or 6, has a primitive prime factor that divides it to an odd power.

Felipe refers to my first ever paper (mathscinet.ams.org/mathscinet-getitem?mr=789713) from 1985 ! However I have a more recent paper that gives a better result along the lines asked for (mathscinet.ams.org/mathscinet-getitem?mr=2997580) which shows that every $2^n-1$, with $n\ne1$ or $6$, has a primitive prime factor that divides it to an odd power.

Source Link
Andrew Granville
Andrew Granville
  • 907
  • 7
  • 8

Felipe refers to my first ever paper (mathscinet.ams.org/mathscinet-getitem?mr=789713) from 1985 ! However I have a more recent paper that gives a better result along the lines asked for (mathscinet.ams.org/mathscinet-getitem?mr=2997580) which shows that every 2^n-1, with n<>1 or 6, has a primitive prime factor that divides it to an odd power.