x,y
... Using Fermat’s Little Theorem Lemma If aN-1 1 mod N for some a relatively prime to N, then it must hold for at least half the choices of a < N Proof Fix some value of a such that aN-1 1 mod N. Suppose b < N Satisfies the test, i.e., bN-1 1 mod N. Then, (a·b)N-1 aN-1·bN-1 aN-1 1 mod N Le ...
... Using Fermat’s Little Theorem Lemma If aN-1 1 mod N for some a relatively prime to N, then it must hold for at least half the choices of a < N Proof Fix some value of a such that aN-1 1 mod N. Suppose b < N Satisfies the test, i.e., bN-1 1 mod N. Then, (a·b)N-1 aN-1·bN-1 aN-1 1 mod N Le ...
PRIMALITY TESTING A Journey from Fermat to AKS
... gcd(a, n)=1, an−1 ≡ 1 (mod n), a is called a Fermat liar for n, i.e., n is, in this case, a pseudoprime to base a. Now, we state our theorem and prove it. Theorem : Given any composite non-Carmichael number n, the number of Fermat witnesses in the set (1,2,3...(n − 1)) coprime to n is greater than o ...
... gcd(a, n)=1, an−1 ≡ 1 (mod n), a is called a Fermat liar for n, i.e., n is, in this case, a pseudoprime to base a. Now, we state our theorem and prove it. Theorem : Given any composite non-Carmichael number n, the number of Fermat witnesses in the set (1,2,3...(n − 1)) coprime to n is greater than o ...
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Annotations on Divisibility Test
... ⇒ ((d2.10+d1)10+d0) ≡ ((d2.3+d1)3 + d0 (mod 7)) as we know that 10 ≡ 3(mod 7). So, we can find the remainder (mod 7) as follows. Start with the leading digit then repeatedly apply the operation i.e multiply by 3 then add the next digit, doing all of the arithmetic (mod 7). For example, let’s use thi ...
... ⇒ ((d2.10+d1)10+d0) ≡ ((d2.3+d1)3 + d0 (mod 7)) as we know that 10 ≡ 3(mod 7). So, we can find the remainder (mod 7) as follows. Start with the leading digit then repeatedly apply the operation i.e multiply by 3 then add the next digit, doing all of the arithmetic (mod 7). For example, let’s use thi ...
Collatz conjecture
The Collatz conjecture is a conjecture in mathematics named after Lothar Collatz, who first proposed it in 1937. The conjecture is also known as the 3n + 1 conjecture, the Ulam conjecture (after Stanisław Ulam), Kakutani's problem (after Shizuo Kakutani), the Thwaites conjecture (after Sir Bryan Thwaites), Hasse's algorithm (after Helmut Hasse), or the Syracuse problem; the sequence of numbers involved is referred to as the hailstone sequence or hailstone numbers (because the values are usually subject to multiple descents and ascents like hailstones in a cloud), or as wondrous numbers.Take any natural number n. If n is even, divide it by 2 to get n / 2. If n is odd, multiply it by 3 and add 1 to obtain 3n + 1. Repeat the process (which has been called ""Half Or Triple Plus One"", or HOTPO) indefinitely. The conjecture is that no matter what number you start with, you will always eventually reach 1. The property has also been called oneness.Paul Erdős said about the Collatz conjecture: ""Mathematics may not be ready for such problems."" He also offered $500 for its solution.