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The function $z\mapsto \wp (-z)$ has the same periodicity, poles and normalized Laurent behavior as $\wp (z)$. Uniqueness in its definition therefore makes $\wp$ even, so ${\wp }^{\prime }$ is odd. At a nonzero half-period $b$ with $2b\in \Lambda$, periodicity and oddness give ${\wp }^{\prime }(b)={\wp }^{\prime }(-b)=-{\wp }^{\prime }(b)$, hence zero. Modulo $\Lambda$ the three such points are $1/2$, $\tau /2$, $(1+\tau )/2$.

An elliptic function has equal total orders of zeros and poles in a period cell, by the argument principle and cancellation of opposite boundary edges. The derivative has exactly one pole modulo the lattice, of order three at zero. The three displayed zeros therefore exhaust its zeros, each with multiplicity one:

The only potential poles of $h$ outside the lattice are at $z=a$ and $z=-a$. Since $2a\notin \Lambda$, these points are distinct modulo the lattice and ${\wp }^{\prime }(a)\ne 0$. At each point the squared factor $(\wp (z)-\wp (a){)}^2$ vanishes to order two, canceling the double pole of the shifted Weierstrass elliptic function. The derivative term is regular there. Thus $h$ has no poles outside $\Lambda$.

Near zero write $\wp (z)=z^{-2}+c_2{z}^2+O(z^4)$, using evenness and the normalization. Taylor expansion at $a$ givesConsequentlyThe $z^{-3}$ terms cancel exactly, and there are no $z^{-2}$ terms. Thus there is at most one simple pole per period cell. The sum of residues of an elliptic function over a period cell is zero, again because opposite boundary contour integrals cancel; its sole possible residue must vanish. Hence $h$ is entire. Periodicity makes it bounded on the plane, and Liouville's theorem makes it constant. In fact $h$ is odd, so that constant is zero. The resulting Weierstrass shift-difference identity by pole cancellation is

Choose an integer $b$ whose reduction is a primitive root modulo $p$. At least one of $b,b+p$ satisfies $g^{p-1}\not\equiv 1(\mathrm{mod}{p}^2)$, because the binomial theorem givesand the added coefficient is nonzero modulo $p$. Both choices still generate the multiplicative group modulo $p$. Choose such a $g$ and write $g^{p-1}=1+up$ with $p\nmid u$.

For $k\ge 1$ and $p\nmid v$, another binomial expansion showsThe intermediate binomial coefficients are divisible by $p$, and the final term also has valuation at least $k+2$ because $p$ is odd. Induction therefore givesThus $g^{p-1}$ has order exactly $p^{n-1}$ modulo $p^n$. The order of $g$ itself is divisible by $p-1$, by reduction modulo $p$, and divides the number $p^{n-1}(p-1)$ of invertible residue classes. Its order must consequently be precisely that number. This proves the lifting a primitive root to odd prime powers result andFor the requested example choose $g=2$. Modulo $11$, $2^5\equiv -1$ and $2^2\not\equiv 1$, excluding the proper divisors $1,2,5$ of ten, so its order is ten. Also $2^{10}-1=1023=11\cdot 93$, which is not divisible by $11^2$. Hence

Define the multicolour graph Ramsey number $R_k(s_1,\dots ,s_k)$ as the least $n$ such that every colouring of the edges of the complete graph $K_n$ with $k$ colours contains, for some $i$, a complete subgraph on $s_i$ vertices all of whose edges have colour $i$.

Prove finiteness by induction on ${\sum }_i{s}_i$, temporarily allowing $s_i=1$. If some $s_i=1$, a single vertex supplies the required vacuous monochromatic clique, so the Ramsey number is one. Otherwise put $N_i=R_k(s_1,\dots ,s_i-1,\dots ,s_k)$, which is finite by induction, and takeFix a vertex $v$ and partition its other $n-1$ vertices according to the colour of their edge to $v$. Some colour-$i$ class has at least $N_i$ vertices: if every class had at most $N_i-1$, their total would be at most ${\sum }_i{N}_i-k=n-2$. In that class the inductive property either gives a colour-$j$ clique of size $s_j$ with $j\ne i$, or a colour-$i$ clique of size $s_i-1$. In the latter case add $v$. ThusIn particular the requested two-colour diagonal number is $R(s)=R_2(s,s)$, so it exists for every $s\ge 2$.

Now colour each positive integer by its partition class, say there are $k$ classes, and take $m=R_k(3,\dots ,3)$. Colour the edge $ij$, $1\le i<j\le m$, by the class of $j-i$. There is a monochromatic triangle $a<b<c$. Its three differences $x=b-a$, $y=c-b$, $z=c-a$ are positive integers in one class, andThis is the finite-Ramsey proof of the partition conclusion. The statement does not require $x$ and $y$ to be distinct.

For Hartogs theorem, collect all well-order relations on all subsets of a set $x$. They form a set, since the relations belong to the power set of $x\times x$. Replacement collects their ordinal order types in a set $T$. DefineIf there were an injection $\alpha \to x$, transfer the well-order of $\alpha$ to its image. Its type would put $\alpha$ in $T$, giving $\alpha +1\le \alpha$, impossible. Thus some ordinal does not inject into $x$, without using choice.

Assume the axiom of choice and let a nonempty partially ordered set $P$ have an upper bound for every chain. Suppose it has no maximal element. Choose, from every subset of $P$ with a nonempty set of strict upper bounds, one such strict upper bound. Every chain has a strict upper bound: first take an upper bound and then a strictly larger element, using the absence of maximal elements. Take a Hartogs ordinal $\alpha$ for $P$. Transfinite recursion chooses $p_{\beta }$, for each $\beta <\alpha$, strictly above all previously chosen elements. Their predecessors form a chain at every stage, so this recursion is defined and strictly increasing. It gives an injection $\alpha \to P$, contradicting the choice of $\alpha$. Hence $P$ has a maximal element, proving Zorn's lemma.

For the well-ordering theorem, take a choice function on the nonempty subsets of $x$ and recursively choose an element of $x$ not previously selected. If this process had not exhausted $x$ at any stage below its Hartogs ordinal $\alpha$, it would define an injection $\alpha \to x$, again impossible. Therefore it exhausts $x$ at some ordinal stage $\beta <\alpha$. Order each element by its unique selection index; this transfers the ordinal well-order to all of $x$. The empty set is already well-ordered. These proofs use transfinite recursion, not an unstated fixed-point theorem.