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Colour $n>0$ by $(v_2(n)\mathrm{mod}2,\lfloor {\log }_{2}n\rfloor \mathrm{mod}2)$. No increasing infinite sequence can make all its sums $x_i+x_j$ and $x_i+2x_j$, $i<j$, one colour. If its 2-adic valuations are bounded, two terms of equal valuation and equal odd part modulo four give pair sums of different valuation parity. If they are unbounded, a later term divisible by a sufficiently large power of two prevents adding a fixed earlier term from crossing either dyadic boundary; the two resulting sums have adjacent logarithmic scale indices.

There is a finite coloring of the positive integers for which no infinite set has all pairwise sums and pairwise products in one color. Refining this coloring by the parity of the 2-adic valuation also prevents a constant infinite sequence from evading the obstruction.

Write $v_2(m)$ for the 2-adic valuation of a nonzero integer $m$. Then $d(a,b)=2^{-v_2(a-b)}$ for unequal arguments. It is positive for $a\ne b$, vanishes exactly when $a=b$, and is symmetric because $v_2(-m)=v_2(m)$.

For the triangle inequality, if $2^r$ divides $a-b$ and $2^s$ divides $b-c$, then $2^{\min (r,s)}$ divides their sum $a-c$. ThusCoincident arguments also satisfy these inequalities immediately. Consequently $d$ is a metric, and indeed an ultrametric induced by the 2-adic absolute value.

For the sequence, $x_n-(-1)=2^n$, soThe same sequence diverges in the usual Euclidean metric; the convergence here depends on the 2-adic absolute value.

For $a<b$, give the pair $\{a,b\}$ the color determined by $v_2(b-a)\mathrm{mod}2$, where $v_2$ is the 2-adic valuation: use red for even valuation and blue for odd valuation. This finite coloring works because the pairs $\{a,a+d\}$ and $\{a,a+2d\}$ have valuations $v_2(d)$ and $v_2(d)+1$. They therefore have opposite colors. In particular, no three-term arithmetic progression has all its pairs monochromatic.

For the second assertion, $m=1$ is immediate: every singleton is a blue complete arithmetic progression, since it has no pairs to check. Assume $m\ge 2$. If some $m$-term arithmetic progression has all its pairs blue, we are done. Otherwise, every such arithmetic progression contains a red pair. For each $(a,d)\in {\mathbb{N}}^2$, choose, say lexicographically, indices $0\le r<s<m$ for which $\{a+rd,a+sd\}$ is red, and color $(a,d)$ by this chosen pair $(r,s)$. This is a finite coloring with at most $\left(\genfrac{}{}{0ex}{}{m}{2}\right)$ colors.

We need a lattice pattern that works whichever pair becomes the common color. Fix, before applying the Gallai theorem for an integer lattice, the finite setTranslate $F$ into the positive quadrant and apply the Gallai theorem for an integer lattice to that translated pattern. Absorbing the translation into the base point gives integers $a_0,d_0$ and $q>0$ such that $(a_0,d_0)+qF\subseteq {\mathbb{N}}^2$ is monochromatic. Write $(r,s)$ for its common color. Since $(0,0)\in F$, the base point itself has positive coordinates.

For each $0\le i,j<m$, the parameter pointbelongs to that monochromatic copy and has chosen red pair $(r,s)$. Its red endpoints simplify toConsequently, puttinggives two $m$-term arithmetic progressions of positive common difference $q(s-r)$, with every cross-pair red. These arithmetic progressions really are disjoint: the parameter point corresponding to $i=m-1,j=0$ has positive second coordinate, so $d_0>q(m-1)$. HenceThis proves the arithmetic-progression bipartite Ramsey dichotomy, including the required disjointness.

The elliptic-curve discriminant of $y^2=x^3-12x^2+20x$ is $16\cdot 20^2\cdot (144-80)=409600=2^{14}{5}^2$. Thus three and eleven are primes of good reduction. Counting the reduced points gives four points over ${\mathbb{F}}_3$ and sixteen over ${\mathbb{F}}_{11}$. For example, the numbers of affine points above successive $x=0,1,2$ modulo three are $1,1,1$. Modulo eleven, for $x=0,\dots ,10$, they are $1,2,1,2,0,0,2,2,2,2,1$; add the point at infinity in each case.

The reduction of torsion points on an elliptic curve is injective on torsion prime to the residue characteristic. The three-primary rational torsion injects into the group of order sixteen at eleven, so is trivial; likewise the eleven-primary torsion injects into the group of order four at three. Every other primary component injects at both primes. Therefore the rational torsion subgroup has order dividing four. All four rational 2-torsion points are visible, so

The rational point $P=(1,3)$ lies on the curve. The tangent slope in the elliptic-curve addition formula is $(3-24+20)/6=-1/6$, giving $2P=(361/36,-323/216)$. The integral substitution $X=x-4$ changes the equation to $y^2=X^3-28X-48$; the doubled point has $X=217/36$. The Nagell–Lutz theorem says that a rational torsion point on this integral short equation has integral coordinates, so $2P$, and hence $P$, is nontorsion. An infinite-order point is $(1,3)$.

We prove the rank bound by two-isogeny descent. For $a=-12,b=20$, the two-isogeny formula gives the partnerand maps $\varphi :E\to E^{\prime }$, $\psi :E^{\prime }\to E$ with $\psi \varphi =[2]$. The two-torsion square-class homomorphism $\alpha :E(\mathbb{Q})\to {\mathbb{Q}}^{\ast }/({\mathbb{Q}}^{\ast }{)}^2$ is $\alpha (x,y)=[x]$ away from $O,(0,0)$, with values $1$ and $[20]=[5]$ at those exceptional points. Its kernel is $\psi E^{\prime }(\mathbb{Q})$. The corresponding map ${\alpha }^{\prime }$ on $E^{\prime }$ has kernel $\varphi E(\mathbb{Q})$.

The prime-support bound in two-isogeny descent can be checked directly here. For a prime $\ell \nmid 20$, if $v_{\ell }(x)>0$ then $x^2-12x+20$ is a unit, so $2v_{\ell }(y)=v_{\ell }(x)$ is even; if $v_{\ell }(x)<0$, the $x^3$ term has strictly least valuation and $2v_{\ell }(y)=3v_{\ell }(x)$ is even. Thus the square class of $x$ is supported only on two and five. Also $x(x-2)(x-10)\ge 0$ forces $x\in [0,2]\cup [10,\infty )$, so nonzero $x$ is positive. The possible classes are $1,2,5,10$, and the four torsion points provide all of them:Consequently $\#\alpha (E(\mathbb{Q}))=4$.

The same valuation argument on $E^{\prime }$ gives possible signed classes $1,-1,2,-2$, since its coefficient $b^{\prime }=64$ has only the prime two. The identity gives class $1$, and $(-4,8)\in E^{\prime }(\mathbb{Q})$ gives class $-1$. For $d=\pm 2$, the two-isogeny descent quartic iswith integral coprime $U,V$ after scaling a rational solution. If $U$ is odd, its right side is $2$ or $10$ modulo sixteen for $d=2$, and $14$ or $6$ for $d=-2$, according to the parity of $V$. None is a square. If $U=2u$ is even, $V$ is odd and the right side is $32(\pm u^4+3u^2{V}^2\pm V^4)$. The bracket is odd for either parity of $u$, so its 2-adic valuation is five, again impossible for a square. The signed divisor-two obstruction for an isogeny covering excludes both classes. Hence ${\alpha }^{\prime }(E^{\prime }(\mathbb{Q}))=\{1,-1\}$.

For completeness, the index correction in the two-isogeny index formula over a number field is one: $\ker \psi =\{O,(0,0)\}$ and $\varphi ((2,0))=(0,0)$, so the whole kernel lies in $\varphi E(\mathbb{Q})$. ThusThe Mordell-Weil theorem and the four rational 2-torsion points identify this index as $2^{r+2}$. Therefore $2^{r+2}=8$ andThis is the rank-one elliptic curve with roots zero two and ten. We have exhibited a point of infinite order but have not needed to prove that it generates the free part.

For a positive integer $n$, let $v_2(n)$ be its 2-adic valuation and let $k(n)=\lfloor {\log }_{2}n\rfloor$. Use the four-colour dyadic valuation and scale colouringSuppose an increasing infinite sequence had all its two indicated kinds of pair sums in one colour. There are two exhaustive possibilities for its valuations.

If the valuations are bounded, infinitely many terms have one fixed valuation $t$. Among them, infinitely many have the same odd part modulo four. Choose two such terms $a<b$. Their odd parts have sum congruent to two modulo four, while the odd part of $a+2b$ is odd. HenceTheir first colour coordinates differ, contradicting monochromaticity.

If the valuations are unbounded, fix one term $a$ and choose a later $b$ with $2^{v_2(b)}>a$. Put $k=\lfloor {\log }_{2}b\rfloor$. The distance $2^{k+1}-b$ is a positive multiple of $2^{v_2(b)}$, so it exceeds $a$. Similarly $2^{k+2}-2b$ exceeds $a$. Therefore adding $a$ crosses neither upper dyadic boundary:Their second colour coordinates differ, again a contradiction. Thus this four-colouring admits no such increasing infinite sequence.

Rado's criterion. Let $A$ be a matrix over the rational numbers, with columns $a_1,\dots ,a_q$. A partition regular matrix is one for which every finite coloring of the positive integers gives a monochromatic positive vector $x\in {\mathbb{N}}^q$ with $Ax=0$; repeated coordinates are allowed. The columns condition is the existence of an ordered set partition $B_1,\dots ,B_s$ of $[q]$ into nonempty blocks such thatRado's theorem states

Necessity. Clear denominators so the columns are integer vectors. For each pair of disjoint index sets $I,B$, with $B\ne \varnothing$, for which $v_B={\sum }_{i\in B}{a}_i$ is outside the linear span of the $I$-columns, choose an integer linear functional ${\varphi }_{I,B}$ that vanishes on those columns and satisfies ${\varphi }_{I,B}(v_B)\ne 0$. Such a functional exists by extending a basis of the span to one also containing $v_B$, defining a functional on that basis, and clearing its rational denominators.

There are only finitely many such pairs. Choose a prime number $p$ dividing none of these nonzero integer evaluations. If there are no such pairs, any prime will do. Use the last nonzero digit coloring: write $x=p^{v_p(x)}u$, with $p\nmid u$, and color $x$ by $u\mathrm{mod}p\in \{1,\dots ,p-1\}$. Since $A$ is a partition regular matrix, this coloring supplies a monochromatic positive solution, so all its unit parts have a common nonzero residue $d$.

Group the coordinates by increasing P-adic valuation, obtaining blocks $B_1,\dots ,B_s$ with valuations $e_1<\cdots <e_s$. Suppose for some $j$ the sum $v_{B_j}$ lay outside the linear span of the earlier columns, with earlier index set $I$. Apply ${\varphi }_{I,B_j}$ to $Ax=0$. Earlier terms vanish exactly. Divide the remaining integer equation by $p^{e_j}$ and reduce modulo $p$. Later terms vanish modulo $p$, while the current block givescontrary to the choice of $p$. Thus every block sum lies in the earlier linear span, and for the first block this means it is zero. This proves the columns condition by the finite separating-functional proof of the columns condition.

Sufficiency. Suppose the columns condition holds. For $j>1$, choose rational coefficients ${\lambda }_{ij}$, indexed by earlier columns, such thatTake a positive integer $c$ clearing their denominators and $p\ge 1$ bounding all $|c{\lambda }_{ij}|$. We use the allowed monochromatic m-p-c set theorem with $s$ levels. Its M-p-c set is the full setwhere the generators make every displayed expression positive. Positivity is part of this convention, rather than a rule that discards nonpositive expressions. For $i\in B_h$, setAll these numbers lie in the monochromatic set $D$. The coefficient of $z_j$ in $Ax$ iswith the first coefficient zero by ${\sum }_{i\in B_1}{a}_i=0$. Hence $Ax=0$. This Rado solution inside an m-p-c set proves sufficiency.

The equality-or-inequality dichotomy. If every finite coloring gives a monochromatic solution with $x_1=x_2$, the first alternative holds. Otherwise fix a finite coloring $\chi$ admitting no such solution. Given any other finite coloring $\psi$, use the refinement of a finite coloring $n\mapsto (\chi (n),\psi (n))$. Since $A$ is a partition regular matrix, it supplies a solution monochromatic for both colorings. Its first two coordinates cannot be equal, by the defining property of $\chi$, so it is the required unequal solution for $\psi$. Thus one of the two universal alternatives always holds. This is the partition regularity dichotomy under an extra constraint; the alternatives are not asserted to be mutually exclusive.

Both restrictions can fail. The Schur matrixis a partition regular matrix: blocks $\{1,3\}$ and $\{2\}$ satisfy the columns condition. Color $n$ by $v_2(n)\mathrm{mod}2$, using the 2-adic valuation. An equal-coordinate solution would have $x_3=2x_1$, and $v_2(2x_1)=v_2(x_1)+1$, so its coordinates cannot be monochromatic. This exhibits failure of the equality restriction.

Conversely, $A=(1-1)$ satisfies the columns condition in one block. Every solution has $x_1=x_2$, so even the constant one-color finite coloring admits no unequal-coordinate solution. Consequently

No. It is enough to take the prime number $t=2$. By the monochromatic sums-and-products obstruction, there is a finite coloring $\chi$ of $\mathbb{N}$ for which no infinite set has all its pairwise sums and pairwise products in one color. Refine $\chi$ by also recording the parity of the 2-adic valuation.

If a sequence $(x_i)$ made both requested families monochromatic, put $y_i=x_i^2$. If the set of distinct $y_i$ were infinite, an injective subsequence would make all pairwise sums $y_i+y_j$ and products $y_i{y}_j$ monochromatic under $\chi$, a contradiction. Otherwise some $y=x_i^2$ occurs infinitely often. Two occurrences give the sum $2y$ and the product $y^2$, butbecause $v_2(y)$ is even. The refining colors differ, another contradiction.

The two-isogeny descent quartics associated with square classes $2$ and $-2$ on $Y^2=X^3+24X^2+64X$ have no rational points. A rational solution can be scaled to integral $W,U,V$ with coprime integers $U,V$. If $U$ is odd, reduction modulo sixteen gives $2$ or $10$ for the positive class, and $14$ or $6$ for the negative class; none is a quadratic residue. If $U=2u$ is even, then $V$ is odd and the right side is $32(\pm u^4+3u^2{V}^2\pm V^4)$, with the same sign on the two signed terms. The bracket is odd, so its 2-adic valuation is exactly five, impossible for a nonzero square. These two parity cases exhaust primitive solutions.