Timeline for What is wrong with the argument that zero permanent is polynomial?

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Apr 10, 2023 at 14:32	comment	added	Alejandro Quinche		Just to add to your comment user knew that $x \not = 0$ doesn't imply $ x \not = 0 \mod Q$ however he thought that $\|x\|$ was bounded strictly by $Q$ and in that case we have that implication
Apr 10, 2023 at 12:37	comment	added	Alejandro Quinche		I see, obviously this breaks the $NP=P$ argument but if we do get $perm(B)=0$ we would have the implication that $perm(A)=0$ right? I also find funny that for "false positives" we would precisely fall in multiples of $Q$ but maybe there's a simple divisibility argument to be made to make this obvious.
Apr 10, 2023 at 4:59	comment	added	Zach Teitler		I think the point is: you can check if $\operatorname{perm}(B) \neq 0$. If that happens, then OP wanted to deduce that therefore $\operatorname{perm}(B) \not\equiv 0 \pmod{Q}$, hence $\operatorname{perm}(A) \not\equiv 0 \pmod{Q}$, hence $\operatorname{perm}(A) \neq 0$. But the very first deduction is wrong: $x \neq 0$ doesn't imply $x \not\equiv 0 \pmod{Q}$. In conclusion, $\operatorname{perm}(B) \neq 0$ ("easy" to test) doesn't imply $\operatorname{perm}(A) \neq 0$ (what we want to test) (Emil stated the contrapositive, $\operatorname{perm}(A) = 0$ doesn't imply $\operatorname{perm}(B) = 0$.)
Apr 10, 2023 at 4:10	review	Late answers
Apr 10, 2023 at 5:09
S Apr 10, 2023 at 3:53	review	First answers
Apr 10, 2023 at 6:21
S Apr 10, 2023 at 3:53	history	answered	Alejandro Quinche	CC BY-SA 4.0