There exists a concrete polynomial [math] P in mathbb{Z}[x_1,ldots...

lmao

Any reason why it has to be 9 variables and not 8 or 7?

Yes, Con(ZF) implies there aren't any roots. But then there also exists a polynomial for ZF+Con(ZF); for that you need Con(Con(ZF)). And so forth, up the large cardinal hierarchy if you like. Who really knows what the true model of arithmetic looks like. Does it contain no proof of the inconsistency of a Berkeley cardinal? We don't even have intuition for this.

Just a matter of the encoding schema.

Well that's not really surprising at all.

A recurring theme with mathematics is that almost all solutions involving integers only are impossibly hard to deal with, just look at diophantines.

Yeah no, I like math, but not enough to go graph that

Im perhaps not a genius, like you all are on Veeky Forums (no sarcasm intended)

But thats greek to me. I was never any good in math :(

Read a book.

So what is this concrete polynomial? Why don't you tell us?

The integer coefficients are too big.

I suppose you could have just used
a*b*c*d*e*f*g*h*i
but it would have been less fun.