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Calculus and Beyond Homework Help
Proving Divisibility Using Fermat's Theorem
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[QUOTE="bobby2k, post: 4590700, member: 479769"] Hey Sorry it was a typo yes. I am able to show it easy for the prime numbers. Like seven for example: Eulers theorem gives that a^15=a^7*a^7*a = a*a*a=a^3 (mod 7) So a^15-a^3 = 0 (mod 7).The problem comes in 8 and 9. Since we have that phi(8)=4. We have that a^15 = a^5*a^5*a^5=a*a*a=a^3(mod 8). So a^15-a^3 = 0 (mod 8). However in order for what I did to be legal for 8 and 9, I have to have that gcd(a^15-a^3, 8)=1(and the same for 9). If not, what I did in the last paragraph would not be legal. If this is not the case, Eulers theorem can not be used, correct? So i have to show that gcd(a^15-a^3,8)=1 and gcd(a^15-a^3,9)=1? [/QUOTE]
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Proving Divisibility Using Fermat's Theorem
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