For prime [math]n[/math] we have [math]n\ge2,[/math] hence [math]n^2+2\ge6.[/math]
If [math]n[/math] is different from [math]3,[/math] then [math]n^2\equiv1\pmod{3}[/math] by Euler-Fermat, so [math]n^2+2\equiv0\pmod{3},[/math] which means that [math]n^2+2[/math] is divisible by [math]3,[/math] hence not prime.
For [math]n=3[/math] we have [math]n^2+2=11[/math] that's prime.
Không có nhận xét nào:
Đăng nhận xét