0$ and discriminant $d \ne 0$. For specific integer $n>0$ and prime $p$, let $(n)_p = (n_i)$ denote the base $p$ representation of $n$, as in equation~\eqref {Eq.BaseP}. We begin with the case $p=2$, which cannot be handled with Legendre polynomials due to the powers of 2 in their denominators. \begin {theorem} \label {T.Base2} %-------------------------------------------------------------------------- theorem 1 If $p=2$, then $T_n \equiv b \pmod {p}$. \end {theorem} \begin {proof} In the sum in equation~\eqref {Eq.Basic}, the $k=0$ term is $b^n$; the other terms are $0 \pmod {2}$ because $\binom {2k} {k}$ is even for $k>0$. The theorem follows from $b^n \equiv b \pmod {2}$ for $n>0$. \end {proof} When $p$ is an odd prime there are four cases to consider: whether or not $p \mid b$ and whether or not $p \mid d$. We do not treat the case of $p \nmid b d$ except in the general case treated by Theorem~\ref {T.5}. The following three theorems cover the three remaining cases. The easiest case is $p$ dividing both $b$ and $d$, which we treat first. \begin {theorem} \label {T.2} %-------------------------------------------------------------------------------- theorem 2 If $p$ is an odd prime with $p \mid b$ and $p \mid d$, then $p \mid T_n$. \end {theorem} \begin {proof} Because $d=b^2-4c$, $p \mid b$ and $p \mid d$ imply $p \mid c$. Then, using equation~\eqref {Eq.Basic}, it is easy to see that $p$ divides every term of the sum. Hence, $p \mid T_n$. \end {proof} \begin {theorem} %---------------------------------------------------------------------------------------------- theorem 3 Let $p$ be an odd prime with $p \mid b$ and $p \nmid d$. Then the following are true: \\ \indent (1) if $n_i$ is odd for some $0 \le i \le r$, then $p \mid T_n$, \\ \indent (2) if $n_i$ is even and $n_i \le (p-1)/2$ for all $0 \le i \le r$, then $p \nmid T_n$, \\ \indent (3) if $p \mid T_n$, then either $n_i$ is odd or $n_i > (p-1)/2$ for some $0 \le i \le r$. \end {theorem} \begin {proof} Part 1: Using equation~\eqref {Eq.Basic}, $b$ is a factor in every term of the sum when $n$ is odd. Hence, because $p \mid b$, we obtain $p \mid T_n$ for odd $n$. By congruence~\eqref {Eq.Fundamental}, this divisibility extends to all $n$ such that $n_i$ is odd for some $0 \le i \le r$. Part 2: If $n_i$ is even for all $i$, then $n$ is even. For even $n$, the last term of equation~\eqref {Eq.Basic} is $\binom {2n} {n} c^{n/2}$; all the other terms are divisible by $p$ because those terms have a factor of $b$, which $p$ divides. Note that $d=b^2-4c$, $p \mid b$, and $p \nmid d$ imply $p \nmid c$. Because we assume $n_i \le (p-1)/2$ for all $i$, there are no carries when $n$ is added to $n$ in base $p$. Hence, by Kummer's theorem, $ p \nmid \binom {2n} {n}$. Thus, $p$ does not divide $\binom {2n} {n} c^{n/2}$ and we obtain $p \nmid T_n$. Part 3: this is just the contrapositive of part 2. \end {proof} \begin {theorem} \label {T.4} %-------------------------------------------------------------------------------- theorem 4 Let $p$ be an odd prime with $p \nmid b$ and $p \mid d$, then $p \mid T_n$ if and only if $(p+1)/2 \le n_i \le p-1$ for some $0 \le i \le r$. \end {theorem} \begin {proof} From congruence~\eqref {Eq.Fundamental}, we see that $p \mid T_n$ iff $p \mid T_{n_i}$ for some $0 \le i \le r$. Hence, we assume that $n

0$. Using Theorem~\ref {T.Base2}, we conclude $p \mid T_n$ if and only if $p \mid T_1$. For odd prime $p$, the theorem follows from congruence~\eqref {Eq.Fundamental}. \end {proof} \indent For the case $a=b=c=1$, Deutsch and Sagan \cite [Conjecture 5.8] {DS} and David W. Wilson conjecture that, for $n

0$, and Fermat's little theorem. \end {proof} \begin {theorem} \label {T.8} %--------------------------------------------------------------------------------- theorem 8 For odd prime $p$, \begin {equation*} T_{p-1} \equiv d^{(p-1)/2} \; P_{p-1} \bigg ( \frac {b} {\sqrt {d}} \bigg ) \equiv \bigg (\frac {d} {p} \bigg ) \pmod {p}, \end {equation*} where $ (\frac {d} {p} )$ is the Legendre symbol. \end {theorem} \begin {proof} The first congruence is equation~\eqref {Eq.Legendre}. By congruence~\eqref {Eq.Holt2} and the fact that $T_0 = 1$, we obtain $T_{p-1} \equiv d^{(p-1)/2} \; T_0 \equiv d^{(p-1)/2} \pmod {p}$. By Euler's criterion, $(\frac {d} {p} ) \equiv d^{(p-1)/2} \pmod {p}$. Hence, $T_{p-1} \equiv (\frac {d} {p} ) \pmod {p}$. \end {proof} \begin {theorem} %----------------------------------------------------------------------------------------------- theorem 9 For odd prime $p$, \begin {equation*} T_{p-2} \equiv b d^{(p-3)/2} \pmod {p}. \end {equation*} \end {theorem} \begin {proof} If $p=3$, then $T_1=b$. If $p \ge 5$ and $p \mid d$, then clearly $T_{p-2} \equiv 0 \pmod p$ by Theorems~\ref {T.2} and~\ref {T.4}. If $p \nmid d$, then the theorem follows from $T_1=b$ and congruence~\eqref {Eq.Holt2}. \end {proof} \section {Appendix} Here we prove the generalization of the Schur congruence~\eqref {Eq.Schur} and Holt congruence~\eqref {Eq.Holt} for the scaled Legendre polynomials $Q(x,d)$ defined by \eqref {Eq.Q}. In several cases, the parameter $d$ forces the proofs to be somewhat different than the ones by Holt \cite {Holt} and Wahab \cite {Wahab}. In all cases, $p$ is an odd prime and all congruences are modulo $p$. We use $Q_n$ as shorthand for the polynomial $Q_n(x,d)$. As mentioned above, $Q_n$ satisfies the recursion equation \begin {equation} n Q_n = (2n-1)x Q_{n-1} - (n-1)d Q_{n-2} \label {Eq.QRecur} \end {equation} starting with $Q_0=1$. Note that because all the denominators in $Q_n$ are powers of 2, which are relatively prime to $p$, we replace equation~\eqref {Eq.QRecur} by the congruence \begin {equation} n Q_n \equiv (2n-1)x Q_{n-1} - (n-1)d Q_{n-2}. \label {Eq.QRecurMod} \end {equation} We begin by proving four lemmas that are stated, but not proved in Wahab \cite[Lemma 6.2] {Wahab}. The first proof is almost identical to Holt's proof. \begin {lemma} \label {L.1} %-----------------------------------------------------------------------------lemma 1 If $0 \le r