(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. 

$$\begin{array}{l}(Lucascriterion)Iftheprimepdivides{F}_m,wherem\ge 2,then\\ p=k{2}^{(m+2)}+1forsomepositiveintegerk.\\ \\ Proof.\\ \\ Letb=2^{(2^{(m-2)})}(2^{(2^{(m-1)})}-1).Since\\ \\ (1)\\ 2^{2^m}+1\equiv 0\left(modp\right)\\ \\ Wehave\\ \\ (2)\\ b^2=2^{2^{m-1}}\left(2^{2^m}-2\times 2^{2^{m-1}}+1\right)\equiv -2\times 2^{2^m}\\ \equiv -2\times 2^{2^m}+2\left(2^{2^m}+1\right)\equiv 2\left(modp\right)\\ \\ Fromthisitfollowsthat\\ \\ b^{2^{m+1}}\equiv 2^{2^m}\equiv -1\left(modp\right)\\ \\ Andthus\\ \\ b^{2^{m+2}}\equiv 1\left(modp\right)\\ \\ Consequently,accordingtoMultiplicativeOrder,\\ or{d}_{p}b=2^{j}forsomej=m+2.\\ \\ However,ifj<m+2,andsom+1-j=0,then,\\ \\ 0\equiv 1-1={\left(b^{2^j}\right)}^{2^{m+1-\mathrm{j}}}-1=b^{2^{m+1}}-1\equiv 2^{2^m}-1\left(modp\right)\\ \\ \\ Whichcontradictswiththestatement\left(1\right).Hence,\\ \\ (3)\\ or{d}_{p}b=2^{m+2}\\ \\ Thenumberspandbarecoprimeduetostatement(2).\\ Therefore,applyingFerma{t}^{\prime }slittletheorem,\\ i.e.b^{(p-1)}=1(modp),andusingMultiplicativeOrder\\ andstatement(3),weobtain\\ \\ p-1=kor{d}_{p}b=k{2}^{m+2}.\\ \\ Hencep=k{2}^{(m+2)}+1.\\ \\ \\ {}^{\prime }See{\left(Krizek,2001,p.82\right)}^{\prime }\end{array}$$