IMO 2012 N8 (Generalization) #
Find all finite fields $F$ with the following property: for any $r ∈ F$, there exists $a, b ∈ F$ such that $a^2 + b^5 = r$.
Answer #
Any field of cardinality $q ≠ 11$.
Implementation details #
The condition being asked on $F$ is implemented via the predicate good.
For the case $q = 11$ and $q = 31$, we show that good is preserved under field isomorphisms
via good.of_RingEquiv, so we just have to check the fields ZMod 11 and ZMod 31.
For ZMod 11, we show by computer search that $a^2 + b^5 ≠ 7$ for any $a, b ∈ F$.
For ZMod 31, we show that every element of $𝔽_{31}$, other than $22 = 4^2 - 5^5$ and
$27 = 1^2 + 6^5$, takes the form $a^2$, $a^2 + 1$, or $a^2 - 6^5$ for some $a ∈ 𝔽_{31}$.
A finite field F is called good if every element of F
can be written as a^2 + b^5 for some a, b : F.
This is an auxiliary definition, which is not given in the main file since
they aren't as focused on representing elements of F specifically as a^2 + b^5.
Instances For
Classifying good finite fields #
A finite field of cardinality q > 40 is good.
If n is coprime with q - 1, then every element of F is an nth power.
A finite field of cardinality q ≢ 1 (mod 2) is good.
A finite field of cardinality q ≢ 1 (mod 5) is good.
A finite field of cardinality q ≢ 1 (mod 10) is good.
A field of cardinality 11 is not good.
A field of cardinality 31 is good.
Final solution to the generalized version