Past exam of the mathematics course of the University of Cambridge / 2018 / ii / Paper 1 / 1G (Number Theory)

For pairwise coprime integers $m_1,\dots ,m_r$, put $M={\prod }_i{m}_i$. The Chinese remainder theorem states that reduction gives an isomorphismEquivalently, every system $x\equiv a_i(\mathrm{mod}{m}_i)$ has one solution modulo $M$.

For existence, let $M_i=M/m_i$. Since $\gcd (M_i,m_i)=1$, choose $u_i$ with $u_i{M}_i\equiv 1(\mathrm{mod}{m}_i)$. Thenhas every required residue: modulo $m_j$, all terms except the $j$th vanish and the remaining term is $a_j$. For uniqueness, if $x$ and $y$ have the same residues, every $m_i$ divides $x-y$. Pairwise coprimality then makes $M$ divide $x-y$. Thus the simultaneous solution exists and is unique modulo $M$.

Choose distinct primes $p_1,\dots ,p_{1000}$. The moduli $p_k^2$ are pairwise coprime, so the Chinese remainder theorem supplies an integer $N$ satisfyingConsequentlyso $p_k\mid N+k$ but $p_k^2\nmid N+k$. By the definition of a squarefull number, $N+k$ is therefore not squarefull. Adding a sufficiently large multiple of ${\prod }_k{p}_k^2$ makes $N$ positive without changing any congruence. The integers $N+1,\dots ,N+1000$ are the required consecutive integers.