(Lucas’ criterion) If the prime p divides Fm, where m ≥ 2, then p = k 2^(m+2)+1 for some positive integer k.

 Proof that If the prime p divides Fm, where m ≥ 2, then p = k 2^(m+2)+1 for some positive integer k. 

\[\begin{array}{l} (Lucas\;criterion)If\;the\;prime\;p\;divides\;{F_m},where\;m \ge \ 2,then\;\\\ p = k{2^{(m + 2)}} + 1for\;some\;positive\;integer\;k.\\\\ Proof.\\\\ Let\;b = {2^{({2^{(m - 2)}})}}({2^{({2^{(m - 1)}})}} - 1).\;Since\\\\ (1)\\ {2^{{2^m}}} + 1 \equiv 0\;\left( {mod\;\;p} \right)\\\\ We\;{\rm{ }}have\\\\ (2)\\ {b^2} = {2^{{2^{m - 1}}}}\left( {{2^{{2^m}}} - 2 \times {2^{{2^{m - 1}}}} + 1} \right) \equiv - 2 \times {2^{{2^m}}}\\ \equiv - 2 \times {2^{{2^m}}} + 2\left( {{2^{{2^m}}} + 1} \right) \equiv 2\;\;\;\;\left( {\;mod\;\;p} \right)\\\\ From\;{\rm{ }}this\;{\rm{ }}it\;{\rm{ }}follows{\rm{ }}\;that\\\\ {b^{{2^{m + 1}}}} \equiv {2^{{2^m}}} \equiv - 1\;\;\;\left( {mod\;\;p} \right)\\\\ And\;{\rm{ }}thus\\\\ {b^{{2^{m + 2}}}} \equiv 1\;\;\;\left( {mod\;\;p} \right)\\\\ Consequently,\;according\;to\;Multiplicative\;Order,\\\;\;or{d_p}b = {2^j}for\;some\;j = m + 2.\\\\ However,\;\;if\;j\; < m + 2,and\;so\;\;\;m + 1 - j = 0,\;\;then,\\\\ 0 \equiv 1 - 1 = {\left( {{b^{{2^j}}}} \right)^{{2^{m + 1 - {\rm{j}}}}}} - 1 = {b^{{2^{m + 1}}}} - 1\;\;\; \equiv {2^{{2^m}}} - 1\;\;\;\left( {mod\;\;p} \right)\\\\\\ Which{\rm{ }}\;contradicts{\rm{ }}\;with{\rm{ }}\;the{\rm{ }}\;statement{\rm{ }}\left( 1 \right).{\rm{ }}\;Hence,\\\\ (3)\\ or{d_p}b = {2^{m + 2}}\\\\ The\;numbers\;p\;and\;b\;are\;coprime\;due\;to\;statement\;(2).\\ Therefore,\;\;applying\;Fermat's\;little\;theorem,\\\;\;i.e.\;\;{b^{(p - 1)}} = 1(mod\;p),\; and\;using\;Multiplicative\;Order\\\;and\;statement\;(3),we\;obtain\;\\\\ p - 1 = k\;or{d_p}b = \;k{2^{m + 2}}.\\\\ Hence\;\;p = k{2^{(m + 2)}} + 1.\\\\\\ 'See{\rm{ }}\left( {Krizek,{\rm{ }}2001,p.82} \right)' \end{array}\]