# Prime and consecutive numbers

Two distinct primes $$p$$ and $$q$$ are given.

• a) Are there always two consecutive natural numbers such that one of them has the greatest prime divisor $$p$$, and the other has $$q$$?
• b) Prove that there are two such natural numbers if $$q.

a) Let $$p=11$$, $$q=2$$. If $$n$$ is divisible by $$p$$ and $$n+1$$ is divisible by $$q$$, where $$p,q$$ are the greatest prime divisors, then $$n+1$$ is a power of two. Modulo $$11$$, all nonzero remainders are observed for powers of two, that is, the period is $$10$$. This shows that $$2^k-1$$ is a multiple of $$11$$ <=> $$k$$ is a multiple of $$10$$. Thus, for $$n+1=2^k$$ we have $$n=2^k-1$$ is divisible by $$11$$ and $$k$$ is divisible by $$10$$, hence $$2^k-1$$ is divisible by $$2^5-1=31$$, so $$11$$ is not the greatest prime factor.

But I have difficulties with the b)... Can you help?

First, if $$q=2$$, we must have $$p=3$$. This case is trivial, since $$p$$ and $$q$$ are consecutive primes. From now on, assume that $$q>2$$.
Claim: There exist positive integers $$a,b$$ with $$a$$ even, $$b and $$aq-bp:=\varepsilon\in\{\pm 1\}$$.
Proof: Find $$x_0,y_0\in\mathbb{Z}$$ such that $$x_0q-y_0p=1$$. Note that $$(x_0+kp)q-(y_0+kq)p=1$$ for all $$k\in\mathbb{Z}$$. Pick $$k_0$$ such that $$0. If $$x_0+k_0p$$ is even, pick $$a:=x_0+k_0p$$ and $$b:=y_0+k_0q$$. If $$x_0+k_0p$$ is odd, pick $$a=(1-k_0)p-x_0$$ and $$b:=(1-k_0)q-y_0$$. $$\square$$
Now, let $$a,b$$ as in the claim and set $$n:=aq$$. It is trivial that $$q\mid n$$ and $$p\mid n+\varepsilon$$. We have to show that these are the largest prime divisors.
First, it is clear that $$(n+\varepsilon)/p=b, so $$p$$ is the largest prime factor of $$n+\varepsilon$$. Next, $$2q\mid n$$, and $$\frac{n}{2q} = \frac{bp+\varepsilon}{2q}=\frac{p}{2q}\cdot\frac{bp}{p}+\frac{\varepsilon}{2q} so $$n/2q and because $$q>2$$, the largest prime factor of $$n$$ is $$q$$.