Past exam of the mathematics course of the University of Cambridge / 2007 / ia / Paper 4 / 8E (Numbers and Sets)

If the prime $p$ does not divide $x$, then $\gcd (p,x)=1$. The Bezout identity gives integers $a,b$ with $ap+bx=1$. Multiplying by $y$ gives $apy+bxy=y$. If $p\mid xy$, both terms on the left are divisible by $p$, so $p\mid y$. Thus Euclid lemma holds:For completeness, existence in the fundamental theorem of arithmetic follows by strong induction on a positive integer $n>1$. If $n$ is prime there is nothing to prove; otherwise $n=ab$ with $1<a,b<n$, and the inductive prime factorizations of $a,b$ give one for $n$. The integer $1$ has the empty factorization.

For uniqueness, suppose $p_1\cdots p_r=q_1\cdots q_s$ are two prime factorizations. Repeated application of Euclid lemma shows that $p_1$ divides some $q_j$. Both are primes, so they are equal. Reorder the second list, cancel this equal factor, and repeat. All factors must be matched, including their multiplicities. Thus prime factorization exists and is unique up to the order of its factors. For nonzero negative integers one also records the sign.

The Fermat-Euler theorem states that, for a positive integer $q$ and an integer $a$ coprime to $q$, $a^{\phi (q)}\equiv 1(\mathrm{mod}q)$, where the Euler totient function $\phi (q)$ counts residue classes coprime to $q$. The case $q=1$ is trivial. For $q>1$, let $r_1,\dots ,r_{\phi (q)}$ be a reduced residue system. Multiplication by $a$ permutes it: equality of two resulting residues can be cancelled using the modular inverse of $a$, and the products remain coprime to $q$. Multiplying the residues before and after the permutation givesEvery factor on the right is a unit, so its product can be cancelled. This provesThe integer $359$ is prime: none of the primes $2,3,5,7,11,13,17$ up to $\sqrt{359}<19$ divides it. Therefore $\phi (359)=358$. Apply the theorem to $60$ and use the supplied congruence:To connect this to individual digits rather than merely assert periodicity, use long-division remainders $r_0=1$ andInduction gives $r_n\equiv 10^n(\mathrm{mod}359)$. Since $10$ is coprime to $359$, no remainder is zero. The congruence above therefore implies $r_{n+179}=r_n$ for every $n\ge 0$, using the unique representative in $1,\dots ,358$. In particularHenceThis is the mechanism of decimal period from multiplicative order: the period divides $179$.