Let us show that there are infinitely many primes "p\\equiv 1(mod\\ 5)". Let us prove by contraposition. Suppose that there are finitely many such primes, and let these primes be "p_i" for "1\\le i\\le n". Let "m=5p_1p_2...p_n+1". Then "m\\equiv 1(mod\\ 5)" and "m>p_i" for all "1\\le i\\le n". Therefore, "m" is a composite number, and thus "p_s" divides "m" for some "1\\le s\\le n". Consequently, "m=kp_s", that is "p_s(5p_1...p_{s-1}p_{s+1}...p_n)+1=kp_s". It follows that "p_s" divides "kp_s-p_s(5p_1...p_{s-1}p_{s+1}...p_n)=1". It is impossible because of "p_s>1." This contradictions proves that there are infinitely many primes "p\\equiv 1(mod\\ 5)".
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