An ordered pair of integers $(m, n)$ with $1<m<n$ is said to be a Benelux couple if the following two conditions hold : $m$ has the same prime divisors as $n$, and $m+1$ has the same prime divisors as $n+1$.

(a) Find three Benelux couples $(m, n)$ with $m \leqslant 14$.

(b) Prove that there exist infinitely many Benelux couples.

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(a) It is possible to see that $(2,8),(6,48)$ and $(14,224)$ are Benelux couples.

(b) Let $k \geqslant 2$ be an integer and $m=2^{k}-2$. Define $n=m(m+2)=2^{k}\left(2^{k}-2\right)$. Since $m$ is even, $m$ and $n$ have the same prime factors. Also, $n+1=m(m+2)+1=(m+1)^{2}$, so $m+1$ and $n+1$ have the same prime factors, too. We have thus obtained a Benelux couple $\left(2^{k}-2,2^{k}\left(2^{k}-2\right)\right.$ ) for each $k \geqslant 2$.

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