# Prove there exists $2011$ consecutive amazing integers

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 $$n$$.

• Can you confirm that notation $(a,b)$ means $GCD(a,b)$ ? Aug 17, 2019 at 9:24
• Yes, it's \$GCD. Aug 17, 2019 at 9:24
• Have you discovered a "reasonably large" number that isn't amazing?
– user694818
Aug 17, 2019 at 9:26
• No, I have checked only relatively small number. Aug 17, 2019 at 9:28
• I mean anything other than 1 and 2, mostly. I'm trying to imagine what it means for a number to not be amazing.
– user694818
Aug 17, 2019 at 9:31

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.

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