Show that there exist infinitely many $(a,b,c)$ such that $a<b<c$, $LCM(a,b,c)\leq a^2$ and $c \leq a + \sqrt{a}$.

Question

If we pick $p$ such that it is divisor of $a$ bigger or equal than $\sqrt{a}+1$ then $\frac{ap}{p-1}$ is an integer smaller or equal than desired $a+\sqrt{a}$. At first, it was hard for me to find any examples satisfying given conditions, but eventually, it turned out that triplets $(56,60,63)$ and $(72,75,80)$ do, and they both follow the pattern described above. How can I prove that there are infinitely many triplets of that kind?

Answer 1

Consider $$a=(30x+8)(30x+9),b=(30x+7.5)(30x+10),c=(30x+8)(30x+10).$$

We can easily get $a< b< c< a+\sqrt{a}$

If we factorize, we have

$$a=6(15x+4)(10x+3),b=75(4x+1)(3x+1),c=20(15x+4)(3x+1).$$

So, we have

$$\text{lcm}(a,b,c)\le \text{lcm}(6,75,20)(15x+4)(10x+3)(4x+1)(3x+1)=300\times(1800x^4+O(x^3))=540000x^4+O(x^3)$$

Where we have $a^2=810000x^4+O(x^3)$. So for sufficiently large $x$ (actually for all $x\ge 0$) we have $\text{lcm}(a,b,c)\le a^2$