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The function is even, so its Fourier series has $b_n=0$. In the convention $f\sim a_0/2+{\sum }_{n\ge 1}(a_n\cos nx+b_n\sin nx)$,Twice applying integration by parts gives, for $n\ge 1$,Therefore the periodic Fourier representation iswith periodic continuation. The periodic extension is continuous, including the identified endpoints, and piecewise smooth, so the Fourier series converges to it. The coefficients also give absolute convergence and uniform convergence by comparison with $\sum n^{-2}$.

At $x=\pi$, the cosine cancels the alternating coefficient sign; at $x=0$, it is identically one. These evaluations yieldThe first identity solves the Basel problem.

Rouché's theorem. Suppose $f,g$ are holomorphic functions on a neighbourhood of the closure of a bounded domain enclosed by a simple, positively oriented, piecewise smooth curve. If $|g(z)|<|f(z)|$ on the boundary, then $f$ and $f+g$ have the same number of zeros inside, counted with multiplicity.

Apply Rouche theorem with $f(z)=z^4+3$ and $g(z)=e^{iz}$ on the first-quadrant quarter-disc of radius $R=\sqrt{2}$. On the two radial segments, $z^4$ is nonnegative real, so $|f|\ge 3$, whereas $|g|=e^{-\mathrm{Im}z}\le 1$. On the circular arc,At every arc point with positive imaginary part, $|g|<1$, giving the strict inequality; at the remaining endpoint $z=R$, $|f|=7>1=|g|$. Thus the strict boundary inequality holds everywhere, including the origin.

The four zeros of $f$ have modulus $3^{1/4}$ and arguments $\pi /4,3\pi /4,5\pi /4,7\pi /4$. Exactly one lies inside the chosen domain. Rouche theorem gives exactly one zero of $f+g$ there, counted with multiplicity, so that zero is simple. Finally any zero in the open first quadrant satisfiesThere can be no additional zero outside the quarter-disc. Hence

Eisenstein's criterion. Let $f(X)=a_n{X}^n+\cdots +a_0\in \mathbb{Z}[X]$ be a nonconstant primitive polynomial. If a prime number $p$ satisfies $p\nmid a_n$, $p\mid a_j$ for $0\le j<n$, and $p^2\nmid a_0$, then $f$ is an irreducible polynomial over $\mathbb{Q}$.

To prove the Eisenstein criterion, suppose a nontrivial factorization exists over $\mathbb{Q}$. Gauss lemma for polynomials gives $f=gh$ with $g,h\in \mathbb{Z}[X]$ of positive degree. Their leading coefficients are not divisible by $p$, because their product is $a_n$. Reduction modulo $p$ therefore preserves both degrees, andSince the only irreducible factor of the right side is $X$, both reduced factors are nonzero monomials of positive degree. Their constant coefficients are consequently divisible by $p$. This would make $p^2\mid g(0)h(0)=a_0$, a contradiction.

For the requested polynomial, shift the variable by one. The binomial theorem givesIt is monic, all coefficients below the leading one are divisible by $p$, and the constant coefficient is $p$, not divisible by $p^2$. The Eisenstein criterion proves irreducibility of this shifted polynomial. The substitution $X\mapsto X+1$ is an automorphism of $\mathbb{Q}[X]$, with inverse $X\mapsto X-1$, so it preserves nontrivial factorizations. HenceThis is the cyclotomic polynomial for a prime index.