Because the primes are distinct, Fermat's little theorem gives $q^{p-1}\equiv 1(\mathrm{mod}p)$ and $p^{q-1}\equiv 1(\mathrm{mod}q)$. The other term in each modulus vanishes. ThereforeThe distinct primes are coprime, so their product divides this integer. Hence

Apply Fermat's little theorem and Wilson's theorem in the ring of integers modulo n with modulus $p$. They giveA vanishing modular congruence modulo $p$ is precisely divisibility by $p$, so both divisibilities hold, without requiring $p\nmid a$.

Wilson's theorem states that an integer $p>1$ is a prime number if and only ifIn particular this holds for every prime, including $p=2$.

Fermat's little theorem says that for a prime number $p$ and any integer $a$,If $p\nmid a$, cancellation gives the equivalent statement $a^{p-1}\equiv 1(\mathrm{mod}p)$.

Write $g(x)=x^6+x^3+1$. The second polynomial is $g(-x)$. Negating the roots of a polynomial does not change the field generated by them: each original root is minus a new root, and conversely. Hence this splitting field is again $\mathbb{Q}({\zeta }_9)$ and its field extension degree is six. Equivalently, it is the cyclotomic field $\mathbb{Q}({\zeta }_{18})=\mathbb{Q}({\zeta }_9)$, since ${\zeta }_{18}=-{\zeta }_9^5$.

The polynomial identityshows that the roots are the ninth roots of unity which are not third roots. These are precisely the primitive ninth roots of unity, all powers of ${\zeta }_9=e^{2\pi i/9}$. Thus the splitting field is $\mathbb{Q}({\zeta }_9)$. The supplied irreducibility statement makes the displayed degree-six polynomial the minimal polynomial of ${\zeta }_9$, giving

The recessive alternative distinguishes CC from the other two types while assigning CT and TT the same disease-risk category. Pooling those two types therefore removes a distinction that this alternative does not use. The resulting counts are $83,1856$ for cases and $50,1885$ for controls, arranged as CC versus non-CC.

The resulting two-by-two table has $\boxed{(2-1)(2-1)=1}$ reference degree of freedom. Since $8.41>3.84$, reject no association at the five-percent level using the supplied chi-squared test. The observed CC proportion is higher among cases, $83/1939$, than among controls, $50/1935$, so the direction agrees with the proposed increased-risk category.

Use two degrees of freedom from part (iii). Since $8.42>5.99$, the statistic exceeds the supplied five-percent threshold. Thus reject no association at the five-percent level: the observed genotype frequencies provide evidence of a difference between cases and controls. The approximate p-value is $e^{-8.42/2}\approx 0.0148$ under the ${\chi }_2^2$ reference law.

With both column totals fixed, each column is a three-category multinomial sample. Under the null the two columns share one probability vector, which has two free parameters because its three probabilities sum to one. Under the alternative each has its own vector, giving four free parameters. Their difference is two, the reference degrees of freedom for the association test.

Under the null, the common fitted CC frequency is its pooled proportion $133/3874$. Multiplying by the fixed number of cases givesThis estimates the common type frequency from both groups, rather than using the case frequency alone.

For a two-way contingency table, Pearson's statistic isHere $O_{i+},O_{+j}$ are the observed margins and $N$ their grand total. Under the independence null, its usual large-sample reference distribution has $(r-1)(c-1)$ degrees of freedom for an $r$-by-$c$ table.