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Nov 29, 2023 at 4:11 comment added Daniel Weber Intuitively this feels too restricted, perhaps the abc or n conjecture can show something
Nov 28, 2023 at 18:09 history edited Stanley Yao Xiao CC BY-SA 4.0
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Nov 28, 2023 at 17:33 comment added TheBestMagician I like the case $a=b=c=1$, so that $x=y+1$. Because this implies $\gcd(x,y)=1$, we want $\mathrm{lcm}(x_1(x_1+1),x_2(x_2+1),\cdots,x_k(x_k+1))$ to be relatively ``small” compared to the $x_i$. In other words, the gcd of the $x_i(x_i+1)$ needs to be large.
Nov 28, 2023 at 17:18 history asked Stanley Yao Xiao CC BY-SA 4.0