We give a new proof that there are infinitely many primes, relying on van der
Waerden's theorem for coloring the integers, and Fermat's theorem that there
cannot be four squares in an arithmetic progression. We go on to discuss where
else these ideas have come together in the past.Comment: To appear in the American Mathematical Monthl

Granville, Andrew

English

arXiv

We give a new proof that there are infinitely many primes, relying on van der Waerden's theorem for coloring the integers, and Fermat's theorem that there cannot be four squares in an arithmetic progression. We go on to discuss where else these ideas have come together in the past

Granville, A

UCL Discovery

Mathematical Assoc. of America American Mathematical Monthly 121:1 August 23, 2017 11:52 a.m. MonthlyNoteUsingTheirStyleFile.v2.tex page 1Squares in arithmetic progressions andinfinitely many primesAndrew GranvilleAbstract. We give a new proof that there are infinitely many primes, relying on van der Waer-den’s theorem for coloring the integers, and Fermat’s theorem that there cannot be four squaresin an arithmetic progression. We go on to discuss where else these ideas have come togetherin the past.1. INFINITELY MANY PRIMES Levent Alpoge recently gave a rather differentproof [1] that there are infinitely many primes. His starting point was the famous resultof van der Waerden (see, e.g., [9]):van der Waerden’s Theorem. Fix integersm ≥ 2 and ` ≥ 3. If every positive integeris assigned one of m colors, in any way at all, then there is an `-term arithmeticprogression of integers which each have the same color.Using a clever coloring in van der Waerden’s theorem, and some elementary numbertheory, Alpoge deduced that there are infinitely many primes. We proceed from van derWaerden’s theorem a little differently, employing a famous result of Fermat (see, e.g.,[6]):Fermat’s Theorem. There are no four-term arithmetic progressions of distinct integersquares.From these two results we deduce the following:Theorem 1. There are infinitely many primes.Proof. If there are only finitely many primes p1, . . . , pk, then every integer n can bewritten as pe11 · · · pekk for some integers e1, e2, . . . , ek ≥ 0. We can write each of theseexponents ej asej = 2qj + rj, where rj is the “remainder” when dividing ej by 2,and equals 0 or 1. Therefore if we letR = pr11 · · · prkkthen R is a squarefree integer that divides n, andn/R is the square of an integer.(In fact, n/R = Q2 where Q = pq11 · · · pqkk .)We will use 2k colors to color the integers: Integer n is colored by the vector(r1, . . . , rk). By van der Waerden’s theorem there are four integers in arithmetic pro-gressionA,A+D,A+ 2D,A+ 3D, with D ≥ 1,January 2014] PRIMES AND PROGRESSIONS 1Mathematical Assoc. of America American Mathematical Monthly 121:1 August 23, 2017 11:52 a.m. MonthlyNoteUsingTheirStyleFile.v2.tex page 2which all have the same color (r1, . . . , rk). NowR = pr11 · · · prkk divides each of thesenumbers, so also divides D = (A+D)− A. Letting a = A/R and d = D/R, wesee thata, a+ d, a+ 2d, a+ 3d are four squares in arithmetic progression,contradicting Fermat’s theorem.These ideas have come together before to make a rather different, not-too-obviousdeduction:2. THE NUMBER OF SQUARES IN A LONG ARITHMETIC PROGRESSIONLet Q(N) denote the maximum number of squares that there can be in an arithmeticprogression of lengthN . A slight refinement of the Erdo˝s–Rudin conjecture states thatthe maximum number is attained by the arithmetic progression{24n+ 1 : 0 ≤ n ≤ N − 1}which contains√83N squares, plus or minus one. From Fermat’s theorem one easilysees thatQ(N) ≤ 3N + 34,but it is difficult to see how to improve the bound to, say, Q(N) ≤ δN + b for someconstant δ < 34.It was this problem that inspired one of the most influential results [8] in combina-torics and analysis (see, e.g., [5]):Szemere´di’s Theorem. Fix δ > 0 and integer ` ≥ 3. If N is sufficiently large (de-pending on δ and `) then any subset A of {1, 2, . . . , N} with ≥ δN elements, mustcontain an `-term arithmetic progression.van der Waerden’s theorem is a consequence of Szemere´di’s theorem, because if welet δ = 1/m and we color the integers in {1, 2, . . . , N} with m colors, then at leastone of the colors is used for at least N/m integers. We apply Szemere´di’s theorem tothis subset A of {1, 2, . . . , N}, to obtain an `-term arithmetic progression of integerswhich each have the same color.In [7], Szemere´di applied his result to the question of squares in arithmetic progres-sions:Theorem 2 (Szemere´di). For any constant δ > 0, if N is sufficiently large, thenQ(N) < δN .Proof. Suppose that there are at least δN squares in the arithmetic progression {r +ns : n = 1, 2, . . . , N} with s ≥ 1; that is, there exists a subset A of {1, 2, . . . , N}with at least δN elements for whichr + ns is a square, whenever a ∈ A.Szemere´di’s theorem with ` = 4 then implies that A contains a four-term arithmeticprogression, say u+ jv for j = 0, 1, 2, 3. For these values of n, we have r + ns =a+ jd, where a = r + us and d = vs > 1. That is, we have shown thata, a+ d, a+ 2d, a+ 3d are four squares in arithmetic progression,contradicting Fermat’s theorem.2 c© THE MATHEMATICAL ASSOCIATION OF AMERICA [Monthly 121Mathematical Assoc. of America American Mathematical Monthly 121:1 August 23, 2017 11:52 a.m. MonthlyNoteUsingTheirStyleFile.v2.tex page 33. MORE HEAVY MACHINERY One day over lunch, in late 1989, Bombierishowed me a completely different proof of Theorem 2, this time relying on one of themost influential results in algebraic and arithmetic geometry, Faltings’ theorem [3].Faltings’ theorem is not easy to state, requiring a general understanding of an algebraiccurve and its genus. The basic idea is that an equation in two variables with rationalcoefficients has only finitely many rational solutions (that is, solutions in which thetwo variables are rational numbers), unless the equation “boils down to” an equationof degree 1, 2 or 3. To be precise about “boiling down” involves the concept of genus,which is too complicated to explain here (see, e.g., [3]). Here we only need a simpleconsequence of Faltings’ theorem.Corollary to Faltings’ Theorem. Let b1, b2, . . . , bk be distinct integers with k ≥ 5.Then there are only finitely many rational numbers x for which(x+ b1)(x+ b2) · · · (x+ bk) is the square of a rational number.Another proof of Theorem 2. Fix an integer M > 6/δ. Let B(M) be the total num-ber of rational numbers x and integer 6-tuples b1 = 0 < b2 < . . . < b6 ≤M − 1 forwhich (x+ b1)(x+ b2) · · · (x+ b6) is the square of a rational number. Faltings’ the-orem implies thatB(M) is some finite number, as there are only finitely many choicesfor the bj . We let N be any integer ≥M(B(M) + 5).The interval [0, N − 1] is covered by the sub-intervals Ij for j = 0, 1, 2, . . . , k −1, where Ij denotes the interval [jM, (j + 1)M), and kM is the smallest multiple ofM that is greater than N .Let N := {n : 0 ≤ n ≤ N − 1 and a + nd is a square}, where the arithmeticprogression is chosen so that |N | = Q(N). Let Nj = {n ∈ N : n ∈ Ij} for eachinteger j. Let J be the set of integers j for whichNj has six or more elements.Now if n1 < n2 < . . . < n6 all belong toNj , write x = a/d+ n1 and bi = ni −n1 for i = 1, . . . , 6, so thatb1 = 0 < b2 < . . . < b6 ≤M − 1and each x+ bi = a/d+ ni = (a+ nid)/d, which implies that(x+ b1)(x+ b2) · · · (x+ b6) = (a+ n1d)(a+ n2d) · · · (a+ n6d)d6is the square of a rational number.This gives rise to one of the B(M) solutions counted above, and all the solutionscreated in this way are distinct (since given x, d, b1, . . . , b6 we have each a+ njd =d(x+ bj)). Therefore the setNj gives rise to(|Nj |6)such solutions, and so in total wehave ∑j∈J(|Nj|6)≤ B(M).It is easy to verify that r ≤ 5 + (r6)for all integers r ≥ 1, and soQ(N) = |N | =k−1∑j=0|Nj| ≤k−1∑j=05 +∑j∈J(|Nj|6)≤ 5k +B(M),January 2014] PRIMES AND PROGRESSIONS 3Mathematical Assoc. of America American Mathematical Monthly 121:1 August 23, 2017 11:52 a.m. MonthlyNoteUsingTheirStyleFile.v2.tex page 4as |Nj| ≤ 5 if j 6∈ J . Finally, as k ≤ N/M + 1 we haveQ(N) ≤ 5k +B(M) ≤ 5NM+ (B(M) + 5) ≤ 6NM< δN,as desired.Bombieri [2] went on, together with Granville and Pintz, to combine these twoproofs (along with much more arithmetic geometry machinery), to prove thatQ(N) < N cfor any c > 23, for sufficiently large N . Bombieri and Zannier [4] improved this toc > 35with a rather simpler proof. The conjecture that Q(N) behaves more like aconstant times N1/2 remains open.REFERENCES1. L. Alpoge, van der Waerden and the primes, Amer. Math. Monthly 122 (2015), 784–785.2. E. Bombieri, A. Granville, and J. Pintz, Squares in arithmetic progressions, Duke Math. J. 66 (1992),369–385.3. E. Bombieri and W. Gubler, Heights in Diophantine Geometry, New Mathematical Monographs, Vol. 4,Cambridge Univ. Press, Cambridge 2006.4. E. Bombieri and U. Zannier, A note on squares in arithmetic progressions. II, Atti Accad. Naz. Lincei Cl.Sci. Fis. Mat. Natur. Rend. Lincei 13 (2002), 69–75.5. W.T. Gowers, A new proof of Szemere´di’s theorem, Geom. Funct. Anal. 11 (2001), 465–588.6. J.H. Silverman, The arithmetic of elliptic curves, Springer Verlag, New York, 1986.7. E. Szemere´di, The number of squares in an arithmetic progression, Studia Sci. Math. Hungar. 9 (1974),417.8. E. Szemere´di, On sets of integers containing no k elements in arithmetic progression, Acta Arith. 27(1975), 199–245.9. T. Tao and Van Vu, Additive Combinatorics, Cambridge Univ. Press, Cambridge, 2006.ANDREW GRANVILLE De´partement de mathe´matiques et de statistique, Universite´ de Montre´al, CP 6128succ. Centre-Ville, Montre´al, QC H3C 3J7, Canadaandrew@dms.umontreal.caandDepartment of Mathematics, University College London, Gower Street, London, WC1E 6BT, United Kingdoma.granville@ucl.ac.uk4 c© THE MATHEMATICAL ASSOCIATION OF AMERICA [Monthly 121

Squares in arithmetic progressions and infinitely many primes

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