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Clear a common denominator first, so that all coefficients are nonzero integers. This changes neither the solutions nor partition regularity. We prove the one-equation criterion directly; no form of Rado's theorem is used as an assumption.

For necessity, choose a prime number $p>{\sum }_i|a_i|$. Colour a positive integer by its first nonzero base-$p$ digit: writing $x=p^{v_p(x)}u$ with $p\nmid u$, use the colour $u(\mathrm{mod}p)\in \{1,\dots ,p-1\}$. Suppose the equation has a monochromatic solution. Let $v$ be the smallest P-adic valuation of its entries and put $I=\{i:v_p(x_i)=v\}$, a nonempty set. Divide the equation by $p^v$ and reduce modulo $p$. All entries indexed by $I$ contribute the same nonzero digit $u$, and all other terms vanish. HenceIt follows that $p$ divides ${\sum }_{i\in I}{a}_i$. The absolute value of this sum is less than $p$, so $\boxed{\sum _{i\in I}{a}_i=0}$.

For sufficiency, take a nonempty zero-sum index set $I$ and put $S={\sum }_{i\notin I}{a}_i$. If $S=0$, the full coefficient sum is zero and the constant vector $(1,\dots ,1)$ is already a monochromatic solution in every colouring. Otherwise choose $j\in I$, put $t=|a_j|$, and set $b_j=-\mathrm{sgn}(a_j)S$, $b_i=0$ for $i\in I\setminus \{j\}$. ThenAdd the same sufficiently large nonnegative integer $H$ to all the $b_i$, obtaining $c_i=b_i+H\ge 0$. The sum is unchanged because ${\sum }_{i\in I}{a}_i=0$. Let $K={\max }_{i\in I}{c}_i$.

Apply the scaled Brauer progression theorem, proved above from the permitted Van der Waerden theorem, to obtain one-colour numbers $td$ and $a,a+d,\dots ,a+Kd$. DefineAll entries are positive and monochromatic, andTherefore the row is partition regular exactly when some nonempty coefficient subset sums to zero. Repetition of entries is allowed by the definition of partition regularity.

The general modular form, the nonuniform Frankl-Wilson theorem, can be stated as follows. Let $p$ be a prime number, and let $L\subseteq {\mathbb{F}}_p$ have $s$ elements. Suppose $\mathcal{F}\subseteq \mathcal{P}([n])$ has $|A|\mathrm{mod}p\notin L$ for every member, while $|A\cap B|\mathrm{mod}p\in L$ for any distinct members. ThenThe uniform Frankl-Wilson theorem sharpens this to $|\mathcal{F}|\le \left(\genfrac{}{}{0ex}{}{n}{s}\right)$ when all members have size $k$ and $0\le s\le \min \{k,n-k\}$. Both forms require a prime modulus and exclusion of the self-intersection residue.

For the general proof, work over the finite field ${\mathbb{F}}_p$. Associate to each member $A$ the intersection polynomialAt the characteristic vectors of sets $1_B$, this is zero if $B\ne A$, and nonzero if $B=A$. Thus these polynomials, regarded as functions on the Boolean hypercube, are linearly independent: evaluating any relation at $1_A$ isolates its coefficient. Apply multilinear reduction on the Boolean cube, replacing every positive power of a variable by that variable. Values on the Boolean hypercube remain unchanged, and the resulting multilinear polynomials have degree at most $s$. Their ambient space has a monomial basis consisting of $x_S={\prod }_{i\in S}{x}_i$ for $|S|\le s$, with dimension ${\sum }_{j=0}^{\min (s,n)}\left(\genfrac{}{}{0ex}{}{n}{j}\right)$. This proves the general bound, including $s=0$.

For completeness, obtain the uniform sharpening without dividing by factorials in a finite field. For $s\ge 1$ put $a=k\mathrm{mod}p$ and $g(x)={\sum }_i{x}_i-a$. The functions $x_{S}g$ with $|S|<s$ are linearly independent. Indeed, a relation gives $gh=0$ with $\mathrm{deg}h<s$, so $h$ is supported only on weights in $H=\{j\in [0,n]:j\equiv a(\mathrm{mod}p)\}$. With boundary weights $-1,n+1$, this set has a gap of at least $s+1$: either consecutive allowed weights differ by $p\ge s+1$, or its terminal gap does, since $n\ge k+s\ge a+s$. The alternating sum of $h$ over any interval of $s$ free coordinates vanishes by its degree. Across that gap only one endpoint level can contribute, forcing $h$ to vanish there. Delete that level and repeat across the enlarged gap until $H$ is empty. Hence $h=0$, and independence of the square-free monomials gives the claim. Adjoin these functions to the $f_A$. Evaluation at each family vector eliminates the coefficients of $f_A$, since $g$ vanishes there; the claim eliminates all remaining coefficients. Counting dimensions yieldsThe case $s=0$ permits at most one member directly. The gap argument is an instance of the modular layer vanishing lemma.

For the final application, enumerate the distinct sets as $A_1,\dots ,A_m$ and let $v_i=1_{A_i}\in {\mathbb{R}}^n$ be their characteristic vectors of sets. If $m\le 1$, the desired bound is immediate for $n\ge 1$. Otherwise $|A_i|\ge k$, because $A_i$ intersects another member in $k$ points. At most one member can have size $k$: two distinct $k$-sets cannot have intersection size $k$. The Gram matrix of these vectors isFor real coefficients $c_i$,All terms are nonnegative, with $k>0$. If the expression vanishes, every coefficient corresponding to a set of size greater than $k$ is zero. At most one coefficient remains, and the first term forces it to be zero as well. Thus the characteristic vectors of sets are linearly independent in ${\mathbb{R}}^n$, proving the constant-intersection family boundThis proof explicitly covers the possible member of size exactly $k$; assuming all diagonal corrections were strictly positive would miss that case. Positivity of $k$ is essential: when $k=0$, the empty set and all $n$ singleton sets give $n+1$ members.

An additive valuation on a field is a map $v:K^{\times }\to \Gamma$, where $\Gamma$ is a totally ordered abelian group, with $v(xy)=v(x)+v(y)$ and $v(x+y)\ge \min (v(x),v(y))$ whenever $x+y\ne 0$. Set $v(0)=\infty$. We consider nontrivial valuations; if the trivial valuation is allowed, it must be listed separately from the asserted classification. Two valuations are equivalent when their valuation rings agree, or equivalently their ordered value-group images identify in a way compatible with the maps. Real-valued equivalent nontrivial valuations differ by positive scaling.

For a valuation on $\mathbb{Q}$, every integer has nonnegative value, because it is a sum of copies of $1$ or its negative. If all prime numbers had value zero, unique factorization would make the valuation trivial. Thus some prime $p$ has $v(p)>0$. There cannot be two such primes: for distinct $p,q$, Bézout's identity gives $ap+bq=1$ with integers $a,b$, and the valuation inequality would give $v(1)>0$. All other prime values therefore vanish. Factoring the numerator and denominator of a rational number givesThe image is the cyclic ordered group generated by $v(p)>0$, proving equivalence to the P-adic valuation. This is the classification of nontrivial valuations on the rational numbers; an Archimedean absolute value is not an additive valuation satisfying this ultrametric inequality.

The simple-root form of Hensel's lemma says that for a complete discrete valuation ring $\mathcal{O}$ with maximal ideal $\mathfrak{p}$, a polynomial $f\in \mathcal{O}[X]$ and $a\in \mathcal{O}$ with $f(a)\equiv 0(\mathrm{mod}\mathfrak{p})$ and $f^{\prime }(a)\not\equiv 0(\mathrm{mod}\mathfrak{p})$ have a unique root in $a+\mathfrak{p}$.

For $X^3+1$ over ${\mathbb{Q}}_7$, all roots are integral: a negative valuation would make the leading term the unique term of lowest valuation. Modulo $7$, the roots are $3,5,6$, and $3X^2$ is nonzero at each. Hensel's lemma gives three distinct lifts, and a cubic has no further roots. The number is $3$.

For $X^2+2X+4$ over ${\mathbb{Q}}_2$, a root is again integral and reduces to zero modulo $2$. Write it as $2y$ with $y\in {\mathbb{Z}}_2$. The equation becomes $y^2+y+1=0$, impossible modulo $2$. The number is $0$.

For $3X^3+X+3$ over ${\mathbb{Q}}_3$, a putative root of valuation $k<0$ gives term valuations $1+3k,k,1$. The first is uniquely minimal, a contradiction. Thus all roots are integral. Reduction modulo $3$ is $X$, whose root zero is simple since the derivative $9X^2+1$ is a unit everywhere. Hensel's lemma gives exactly one lift. The number is $1$.

For a finite group, the conjugacy classes partition the group, and a class is a singleton exactly when its element belongs to the centre of a group. Since the centre is trivial, the class equation becomeswhere the ${\mathcal{C}}_j$ are all the nonidentity conjugacy classes. If the prime number $p$ divided every $|{\mathcal{C}}_j|$, reduction modulo $p$ would give $0=1$, because $p\mid |G|$. Therefore at least one of these class sizes is not divisible by $p$. Primality now givesIts size is greater than one because the centre is trivial. This proves the prime-to-p conjugacy class lemma.

The conclusion requires $p$ to be prime. The printed question does not explicitly impose this hypothesis. If arbitrary composite divisors are allowed, take the symmetric group $S_3$ and $p=6$: its centre is trivial, its two nonidentity conjugacy classes have sizes $3$ and $2$, and neither is coprime to $6$. Thus the argument proves the intended prime case and also identifies why the unrestricted literal reading is false.

Write a matrix in the upper unitriangular group as $M=I+A$. Here $A^3=0$, and its only potentially nonzero entry in $A^2$ is $(A^2{)}_{13}=ab$. The binomial theorem therefore gives the unitriangular matrix power formulaFor an odd prime number $p$, both $p$ and $\left(\genfrac{}{}{0ex}{}{p}{2}\right)=p(p-1)/2$ vanish in the finite field ${\mathbb{F}}_p$. Thus $M^p=I$. A nonidentity $M$ has order dividing the prime $p$ (by the same exponent-division argument as in part (i)), and so has order exactly $p$.

For $p=2$, take $a=b=1$ and $x=0$. Then $M^2=I+A^2\ne I$, while $M^4=(I+A^2{)}^2=I$. This element has order $4$, not $2$. Hence every nonidentity element has order $p$ exactly for odd $p$. The counterexample at $p=2$ accounts for the quadratic term in the matrix power formula.

We first prove the needed instance of Cauchy theorem for groups, rather than invoke it. Let a prime number $q$ divide $|G|$ and consider the setThere are $|G{|}^{q-1}$ such tuples: choose the first $q-1$ entries freely, and the last is forced. Cyclic rotation preserves this set, because moving the first entry to the end conjugates the product, which remains $1$. Each rotation orbit of a group action has either one or $q$ elements. Indeed a nontrivial shift fixing a tuple generates all shifts, since its step has an inverse modulo the prime $q$. The fixed tuples are precisely $(g,\dots ,g)$ with $g^q=1$.

Counting nonfixed orbits in multiples of $q$, and using $q\mid |G{|}^{q-1}$, shows that the number of solutions of $g^q=1$ is divisible by $q$. The identity supplies one solution, so there is a nonidentity solution. If its order of a group element is $d$, divide $q$ by $d$: the remainder would give a smaller positive exponent producing the identity, so $d\mid q$. Hence $d=q$. This proves the required prime-order existence result by the cyclic-tuple proof of Cauchy theorem.

Our hypothesis says that this nonidentity element has order $p$, so $q=p$. Every prime divisor of $|G|$ is therefore $p$, which givesThus a finite group of prime exponent has prime-power order. The case $n=0$ includes the trivial group. No group-theoretic counting theorem was assumed in the tuple argument.