1.5.09 Find a multiplicative inverse to n mod n^2 + 1
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Determine the inverse modulo m of some (relatively prime) integer n. Thus, find the inverse of 16 mod 101 (i.e. an integer c such that 16c ≡ 1 (mod 101)). (a) Perform Euclid's algorithm on 16 and 101. (b) Run Euclid's algorithm backwards to write 1 = 101s + 16t for suitable integers s, t. (c) From the equation 101s + 16t ≡ 1 (mod 101), the multiple of 101 becomes zero (because we are considering congruence) and so we get 16t ≡ 1 (mod 101). Hence the multiplicative inverse of 16 mod 101 is .
Keondre P.
Given n ∈ ℕ, let [a]n denote the equivalence class of a under the relation ‘‘congruence modulo n’’ on the integers. We define the multiplicative inverse of [a]n to be the equivalence class [b]n such that [a]n[b]n = [1]n, if such an equivalence class exists. (Multiplicative inverses are nice because they allow us to perform ‘‘dividing by [a]n’’ by multiplying by the multiplicative inverse of [a]n). (a) Write down the multiplication table for the equivalence classes of the relation ‘‘congruence modulo 5’’. Show that every equivalence class [k]5, where k ≢ 0 (mod 5), has a multiplicative inverse. (b) Prove that if n ∈ ℕ is composite, then there exists an equivalence class [a]n that does not have a multiplicative inverse.
Sri K.
Let n = 3837523. Using a quadratic sieve method, suppose that the following congruences have been found (all mod n): 3397^2 ≡ 2^5 ∙ 5 ∙ 13^2 8077^2 ≡ 2 ∙ 19 9207^2 ≡ 7^4 ∙ 11 ∙ 13 9266^2 ≡ 2 ∙ 3^2 ∙ 5^3 ∙ 7^2 ∙ 13 9398^2 ≡ 5^5 ∙ 19 9820^2 ≡ 3^2 ∙ 5^2 ∙ 13^3 19095^2 ≡ 2^2 ∙ 5 ∙ 11 ∙ 13 ∙ 19 Find a suitable choice of some of these congruences that you can multiply together to get numbers x, y such that x^2 ≡ y^2 (mod n) but x ≢ ±y (mod n), and then use them to factor n. (Only certain choices will be successful, so you might have to try more than one.)
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