SAUDI ARABIAN MATHEMATICAL COMPETITIONS

The given problem can be generalized as follows: Given three positive integers $k,m,n$. Let $k_1,k_2,\dots ,k_m$ be any positive integer. Prove that there are infinitely many pair of positive integers $(a,b)$ such that $$\prod _{k=1}^m\left(a+k_i\right)\mid b(b+k)\text{ but }\prod _{k=1}^m\left(a+k_i\right)\nmid b\text{ and }\prod _{k=1}^m\left(a+k_i\right)\nmid b+k.$$ Proof. Let $k<p_1<p_2<\dots <p_m$ be any $m$ distinct primes and denote $$M=\prod _{k=1}^m\left(a+k_i\right).$$ By Chinese remainder theorem, there are infinitely many positive integers $a>n$ such that $$a\equiv -k_i(\mathrm{mod}{p}_i),\forall i=1,\dots ,m.$$ Hence $M\equiv 0(\mathrm{mod}{p}_1{p}_2\dots p_m)$. Then we write $$M=p_1^{{\alpha }_1}{p}_2^{{\alpha }_2}\dots p_m^{{\alpha }_m}{q}_1^{{\beta }_1}{q}_2^{{\beta }_2}\dots q_s^{{\beta }_s}$$ with ${\alpha }_i\ge 1,i=1,\dots ,m;{\beta }_j\ge 1,j=1,\dots ,s$ and $q_1,q_2,\dots ,q_s$ are $s$ prime divisors of $M$ which are different from $p_1,p_2,\dots ,p_m$. By Chinese remainder theorem, there exist infinitely many positive integers $b>m$ such that $$\{\begin{array}{l}b\equiv 0(\mathrm{mod}\frac{M}{p_m^{{\alpha }_m}})\\ b\equiv -k(\mathrm{mod}{p}_m^{{\alpha }_m})\end{array}$$ This implies that $$M\nmid b,M\nmid b+k\text{ and }M\mid b(b+k).$$ $\square$