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Exercise 2199Number theory › PrimesM1Added 3 Nov 2026

Proth's theorem

#Proth#primes#primality

Statement

State Proth's theorem and apply it to N=3⋅24+1=49N=3\cdot2^4+1=49.

Hint

Proth numbers are k⋅2n+1k\cdot2^n+1 with k<2nk<2^n odd. NN is prime iff there exists aa with a(N−1)/2≡−1(modN)a^{(N-1)/2}\equiv-1\pmod N.

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Solution

Theorem. Let N=k⋅2n+1N=k\cdot2^n+1 with k<2nk<2^n odd. NN is prime iff there exists aa with a(N−1)/2≡−1(modN)a^{(N-1)/2}\equiv-1\pmod N.

Apply to N=49N=49. 49=3⋅24+149=3\cdot2^4+1, so k=3k=3, n=4n=4, (N−1)/2=24(N-1)/2=24. Need a24≡−1(mod49)a^{24}\equiv-1\pmod{49}. Try a=3a=3: 324=(36)4=7294≡729≡−8(mod49)3^{24}=(3^6)^4=729^4\equiv729\equiv-8\pmod{49}... compute more carefully. Actually 49=7249=7^2, not prime, so no such aa exists.

Correction. 49=7249=7^2 is composite; Proth's theorem confirms no such aa exists.

Conclusion. The theorem is used to certify primality of Proth numbers efficiently.