arXiv2026
We prove the irreducibility conjecture for Legendre polynomials proposed by Stieltjes: for every integer $j\geq 1$, both $P_{2j}(x)$ and $P_{2j+1}(x)/x$ are irreducible over the rational numbers. The proof proceeds by contradiction. Starting from a hypothetical nontrivial factorization, we remove the zero root from $P_n$ when present, multiply by a suitable constant, and express the resulting polynomial as a product of two integral polynomials $A(x^2)$ and $B(x^2)$. We then construct the resultant and its odd part, $R=\operatorname{Res}_t(A,B), \mathcal{R}=\frac{|R|}{2^{v_2(R)}}.$ Put $k=\operatorname{deg} A\leq\operatorname{deg} B$ and $m=\lfloor n/2\rfloor$. Using the classical auxiliary polynomial $U_n$ and the differential identity $(1-x^2)(P_n'U_n-P_nU_n')=1-P_n^2,$ together with orthogonality, divisibility properties of the coefficients, and a least-common-multiple estimate, we obtain an upper bound for $\mathcal{R}$. On the other hand, the Legendre differential equation gives a lower bound for the absolute value of the derivative of $AB$ at each root of $A$. We use Chebyshev polynomials and Hadamard's inequality to bound the product of the squared pairwise differences of these roots, and combine this with estimates for the leading coefficients and the power of $2$ in $R$ to obtain a lower bound for $\mathcal{R}$. These estimates yield $m\log4-\log\frac{4(m+1)^2}{\sqrt m} <\frac{\log\mathcal R}{k} <(m+\sqrt{2n-1})\log3 (n\ge64).$ For every $n\ge256$, the lower bound strictly exceeds the upper bound, giving a contradiction. Combining this with established irreducibility results and explicit integer comparisons for the remaining degrees proves irreducibility in every degree.