Prove there exists $2011$ consecutive amazing integers

Question

Recently, I have found this problem:

We call a positive integer $n$ amazing if there exists positive integers $a, b, c$ such that the equality $$n = (b, c)(a, bc) + (c, a)(b, ca) + (a, b)(c, ab)$$ holds. Prove that there exists $2011$ consecutive positive integers which are amazing.

Here some amazing numbers:

In the picture from left to right you find the numbers: $n$, $a$, $b$ and $c$.

I have tried to solve this problem in a lot of different ways, for example using the definition of $GCD$, or divisibility but I can't go on. Any idea?

Note:by $(m, n)$ we denote the greatest common divisor of positive integers $m$ and $n$.

Answer 1

Note that if $n=d^2k$, with $d+2|k$, then with $c=d$, $b=\frac{dk}{d+2}$, then $a=bc$, $(a,b)(c,ab)=bc=d^2\frac{k}{d+2}$, $(b,c)(a,bc)=d^3\frac{k}{d+2}$, and $(c,a)(b,ac)=bc=d^2\frac{k}{d+2}$, so the sum is $n$, which is thus amazing.

So consider a sequence $\delta_1 \geq 6$ and $\delta_{i+1}=\prod_{k=1}^i{(\delta_k^2-1)}$, then, for all $1 \leq i < j$, $(\delta_i-1)^2(\delta_i+1)$ and $(\delta_j-1)^2(\delta_j+1)$ are coprime.

Define $d_i=\delta_i-1$, $P_i=d_i^2(d_i+2)$, where the $P_i$ are pairwise coprime.

By CRT there is some $n+1$ such that for all $1 \leq i \leq 2011$, $n+i$ is divisible by $P_i$, so is amazing.

Answer 2

Claim; For any $j \in \mathbb{N}$ multiples of $j^3+2j^2$ are amazing.

Proof; Set $a=j, b=jk, c =j^2 k$;

One has that $(b, c)(a, bc) + (c, a)(b, ca) + (a, b)(c, ab) = k(j^3+2j^2)$.

Now construct a sequence $c_d, 1 \leq d \leq 2011, c_d \in \mathbb{N}$ with each $c_d$ of the form $j^3+2j^2$ also with the condition that $gcd(c_u,c_v) = 1$ for $u \neq v$.

All that is remaining to do now is to solve the following congruence;

$m \equiv 0$ $ mod $ $c_1$

$m +1 \equiv 0$ $mod$ $c_2$

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$m +2010 \equiv 0$ $mod$ $c_{2011}$