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Can anyone help me? I know that first part is never an integer and second is, so how is possible that the number which is not an integer is divisible by an integer? p is prime number and p>2

lightning777
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  • Have you tried looking at the binomial expansion so see if you can cancel anything? – Eleven-Eleven Nov 12 '16 at 16:42
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    $(2+\sqrt{5})^p = a+b\sqrt{5}$, for $a,b$ integers. There's no way to cancel with $2^{p+1}$ to get an integer. Probably this is a mistake or is a division in another sense as commented by Peter. – diff_math Nov 12 '16 at 16:46
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    What ring are we working in? Should $\sqrt 5$ perhaps be understood as the square root mod $p$? – sbares Nov 12 '16 at 16:49
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    Since diff_math's idea (and mine) does not seem to work, SBareS' idea might be what is meant. But what, if the prime is, lets say $17$ ? Then, $5$ is no quadratic residue, so has no "square-root" – Peter Nov 12 '16 at 16:53
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    In what context did you find this problem? – Eric Towers Nov 12 '16 at 17:20
  • If $p \equiv \pm 1 \pmod 5,$ then $5$ is a quadratic residue, there is some number $s$ with $s^2 \equiv 5 \pmod p.$ – Will Jagy Nov 12 '16 at 17:20
  • @Will But $\ {\rm mod}\ p!:\ 0 \equiv (\color{}{2!+!s})^{\large p}! - 2(2^{\large p}) \equiv 2+s-4,\Rightarrow, s\equiv 2,\Rightarrow, 4\equiv s^2\equiv 5,\Rightarrow,p\mid 1\ \Rightarrow!\Leftarrow$ $\qquad\quad $ – Bill Dubuque Nov 12 '16 at 22:27
  • @BillDubuque sure. I did not go into the problem itself, i was just suggesting a possible way to interpret $\sqrt 5;$ as you say, that does not work either. Given the lack of response from the OP, this is just another problem that was incorrectly paraphrased for this site. – Will Jagy Nov 12 '16 at 22:32

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