Chaladus, S.

Schinzel, A.

Bulgarian Digital Mathematics Library at IMI-BAS

PLISKASTUDIA MATHEMATICA BULGARICAПЛИСКАБЪЛГАРСКИ МАТЕМАТИЧЕСКИ СТУДИИProvided for non-commercial research and educational use.Not for reproduction, distribution or commercial use.The attached copy is furnished for non-commercial research and education use only. Authors are permitted to post this version of the article to their personal websites or institutional repositories and to share with other researchers in the form of electronic reprints. Other uses, including reproduction and distribution, or selling or licensing copies, or posting to third party websites are prohibited.For further information on Pliska Studia Mathematica Bulgarica visit the website of the journal http://www.math.bas.bg/-pliska/ or contact: Editorial Office Pliska Studia Mathematica Bulgarica Institute of Mathematics and Informatics Bulgarian Academy of Sciences Telephone: (+359-2)9792818, FAX:(+359-2)971-36-49 e-mail: pliska@math.bas.bgA DECOMPOSITION OF INTEGER VECTORS. IIS. CHALADUS, A. SCHINZELIn this paper we shall consider integer vectors n =  [/z,, n2, . . . ,  nk] and write for such vectors: h (n )=m ax | /z,1, L{n)=^n\-\-n\-\- • • • + « * .  One of us has recently proved [3] that for every non-zero vector n f Z *  (£ > 1 )  there is a decomposition: ∏ =  Hp +  t'q, u ,v iZ ,  where p, q£Z*  are linearly independent andA ( p ) A ( q ) ≤ 2 A (n )(* - 2>/(*-1).The exponent (k—2)/(k— l )  cannot be improved (see [2], Remark after Lemma 1). It is natural to ask for the best value of the coefficient. We chall answer this question for k =  3 by proving the following two theorems.T h e o r e m  1. For every non-zero vector n f Z 3 there exist linearly inde­pendent vectors p, q ( Z 3, such that n =  ;/p-|-^q, u, v ^ Z  andh (p) h (q) < ^ - 5 - h (n).T h e o r e m  2. For every e > 0  there exists a non-zero vector n£Z3, such that for all non-zero vectors p , q f Z 3 and all u, fc' fQ n = ^ p  +  ^ q impliesA ( p ) A ( q ) > y / ( —  e) h (∏)∙Originally, in the proof of Theorem 1 some computer calculations were used which were kindly performed by Dr. T. Reginska. We than}* her for the help.The proof of Theorem 1 will be based on geometry of numbers. The inner product of two vectors n, m will be denoted by niti, their exterior product by ∏Xm, the area of a plane domain D by A (D ).L e m m a  1 . Let. a,∙, 6 , be real numbers (/= 1,2,3)a n d M 3 the threeci\ (isminors of order two of the matrix not all equal to 0. The area ofb\ b%the domain H :  | а^х +  Ь/у |≤1 (/=1, 2, 3) equals 2 I M xm 2 1+2 I M tM a 1+2 Iж щ щ . ’if each of the numbers \Mi\, |Af3|, | Af3 1 is less that the sum of the two others, and 4/max{|vW, |, \Л42\, \М3\} otherwise.P r o o f .  We may assume without loss of generality that.Mj I =  absa\ я 2 öi b3> 0 , I |≥| M 2 j =  abs I M i  |≥| Af3 | =  absCl2 ö2 0 3# 1  Яз b\ b3PLISKA Studia mathematica bulgarica, Vol. 11, 1991, p. 15-23.∕The affine transformation a1x + b 1y =  X, a2x-{-b2y ≈  Y transforms the domain H into the domainH ':  | * | ≤ l , m ≤ l ;If \ М г \-{-\М3\>\ M i , the domain H' is obtained from the square \X\≤\, I K | ≤ l  by subtracting two rectangular triangles, symmetric to each other with respect to ( 0 , 0 ), with the vertices16 S. Chaiadus, A. Schinzel∕  1 i r l2 --  I A lo  X /1  M o  4 ∙+  ( K  I M.Ä\ ~ S g n ^ ^—  ^  i M *  I * S g n A f J *Hence,A (  Н Ч - 1  ( l ^ l + ’l ^ s l - l ^ i i ) 8V '  .I M2\\M3\If \M%\-t-\M3\≤L\Mi\9 then H' coincides with the square \X\≤\, \Y\≤\ and Л (Н ' )  =  4. Since А (Н) =  Л (H ') ∕j |, the lemma follows.L e m m a  2. If 0 ≤ a ^ b < l y  then the domainD: I x \ ≤ \ ,  \ y \ ≤ l ,  \ a x  +  b y \ ≤ > \ t x 2-\-y2-\-(ax-\ -by)2 ∙<Lcontains an ellipse E with(1) A (E )> n y J -^ ∙P r o o f .  We takeE: f ( x ,  y ) =  x a +  c ( - f 0 T .x + y ) ° ≤  1.where( 2 ) с ≈  max { - f - ( 6 a +  vl),In order to see that |jc|≤1, I j ' l ≤ l  for (x, y ) (E ,  we notice that by (2)( 3 ) min/ (x, у ) = х я, min/ (x, y ) =  —C-P + T  +  1Moreover, for (л:, у) £ E we have by (2)(4) * + У + ( a x + b y ) ' ≤ ± -  ( ^ - a2% + r l-  x *+  -| - ( 0 » +  l ) ( - ^ T x + ^ ) a)≤ - | -/ (x ,  y ) ≤ - 1 - . ∙If for (л:, .y∏E we had \ax-\-by\>\, it would follow(5) x*+y*< -7 f- » hence, by Cauchy-Schwarz inequality( 6 ) (ax +  b y f  ≤  (aa +  b*)(x*+ y*) <  2 . -±- =  1 ,A decomposition of integer vectors. II 17a contradiction. Thus, for (x, y) £ E we have (7) \ax-\-by |≤1.Finally, A (E ) =  n/\/c and since by (2) c < 4/3, (1) follows.L e m m a  3. Let n£Z 3\ { [ 0 , 0, 0 ]}. The lattice of integer vectors. m £Z 3 such that nm =  0  has a basis a ~ [a it a2, a3], b = [£ lt 6 2> з^]> such that(8) a i “ 2 a 2 a 3 /Zi .Ь г b% ( « 1,  n2, n3) ’ &2 Ьз (∏ l ,  n2, /23) ’Clsb3 Ьгn2(∏1, /z2 ,/z8)P r o o f .  Since na =  nb =  0 and a, b are linearly independent, we haven =  c (a X b )for a certain c (Q .  However, the numbersax a2 Ьг b2Cl% (I3 b2 b3anda3 ax b3 b± are rela­tively prime (see e^  g. [1, p. 53]); hence, the formulae ( 8 ) hold with ±  sign on the right-hand side. Changing if necessary tfie order of a, b, we get the lemma. -L e m m a  4. For every vector n £ Z3 different from [0 ,0 , 0] and [ ± 1 > ± 1 , ± 1 ] for any choice of signs, there exists a vector m £Z 3 such that(9) * mn =  0,( 10)andd o0 < h  (m) <  h (n)I (m)<\/2 A(∏).P r o o f .  Without loss of generality we may assume that ( 1 2 ) 0 ≤ n 1≤ n 3≤,n3>0.If n2= n 3 we take[ 1 , 0 , 0 ] if n1 =  0,1 ] if ЯхфО,and we find (9 ) - ( l l )  satisfied, unless n1 =  tio =  n3=\. Therefore, we may assume besides ( 1 2 ) that n2< n 3. (In virtue of Lemma 2  the domainm_ m ,  o, 1 1 0 . 1 ,D: |A|≤1, \V\≤\, и X + ^ - Y | ≤ 1 , X * + r 2+ { ^ r  X + - ^ - Y ^ ≤ ^ -% пз лз rii ^contains an ellipse E with А (Е)>тг \/3/4.Let a, b be a basis, the existence of which is asserted by Lemma 3. The substitutionX —aix + b*y ' 4 3 ПзV = a2x + b 2yv:transforms D into the domain D': \a ,x+b ,y\≤^4j- n3 (/ =  1 ,2 ,3 ) , ∑ (a tx + bi _y)a≤ 2 tf3.18 S. Chaladus, A. SchinzelHence, D' contains an ellipse E' withA (  E ' ) = - | - « 8a \d n  b n—1A ( Е ) > т с (nv nit n3) ≥ nby ( 8 ). Since the packing constant for ellipses is я/v 12, it follows that E' and, hence, D' contains in its interior a point (лг0, y0)^ Z 2 different from (0, 0). Putting ш =  х 0а-Нз;0Ь, we get the assertion of the Lemma.L e m m a  5. If O ≤ a ≤ l ,  0 ≤ 6 ≤ 1  and a-\-b> 1, the area of the hexagon I jcI<  1 ∙ \y |≤1, \ax-\-by\≤\ is greater than [24/(a2 4-6a4- 1)]1/2.P r o o f .  In virtue of Lemma 1 the area in question equals(2ab 4 - 2a -h 2b—a2—b2— 1 )/a b,thus, it remains to prove that for ( a, b) in the domainG: O ≤ a ≤ l ,  O ≤ b ≤  1 , a +  b >  1the following inequality holds/(a, b) =  (2ab  +  2a 4- 2£ -  a2 -  6 2 - 1)2(a2 4 - b2-j- 1 ) - 24a2b2>0 .We have ∂G ≈L x  U La U L3, whereLx =  {(a, 1): O ≤ a ≤ l } ,  L 2 =  { ( L  b): O ≤ b ≤ l } ,  L 3 = { (a ,  1—a): O ≤ a ≤ l } .We find /(a, l )  =  a2 (a — l ) 3(a —5)4-3a2, but for a ≤  1 a » (a — l)? (a— 5 )≥0 , hence /(a, l ) ≥ 3 a 2≥ 0 .  In view of symmetry between a and b, /(1, b)>3b2≥ 0 .  Moreover,/(a, 1 —a) =  8 a2 (1 —a ) 2 (2a— 1)2≥ 0 .  Hence, for (a, 6 ) £∂G we have / (a, b) ≥ 0 with the equality attained only if (а,Ь)ф G It suffices to show that in the interior of G the function /(a, b) has no local extremum.Indeed, putting g (a , b) =  2ab-{-2a—a2 — b2 — 1, we find=  2ag3 +  2 (26 +  2 — 2a)(a2 H- 6 9 +  1) g  —  48aöa, ч ,даOL∂b =  2Ö£ 2 н- 2 (2a +  2 -2 b ) (a 2 +  b2 +  1) g — 48 a2b,hence,a ^ - b ^  =  2 .(a -b ){(a  +  b )g + (a 2 +  b2+ l  ) ( 2  -  2 a -  26)],b § f —a-%-—4(b—ä) [(a-t-ö+ 1 )(a2 +  b2 + 1 ) 1 2 aö (a +  6 )].The. equations ∂f/∂a =  ∂fl∂b =  0 imply a = 6  or(13) (a b) g-\- (a2 b2 1 ) (2 — 2 a —2 ö) =  0 ,( a 4-^4-1 )(a 2 4- £ 2 4- 1) g — 12ab (a +  b)=0.Eliminating g from the above equations we obtain(14) 2 (a2 4- 6 2 4-1 ) [ (# 4 - 6 ) 2 —-1 ] — 12ab (a 4- b)2≈  0. *The left-hand sides of the equations (13) and (14) are symmetric functionsof a, b. Expressing them in terms of s=a-\-b and p =  aby then eliminating /?,we gets (s— 1 )(2s- 1 )(4 s* -  s 4 - 1) =  0.For s =  x + y > l  this is clearly impossible, there remains the possibility a ≈b .  However, in that caseA decomposition of integer vectors. II 19- g j=  16a3—24a2-f- 18a— 4 =  2 ( 2 a — l ) 3 + 3 (2 a  — 1 ) +  1 >  1.L e m m a  6 . For every nonzero vector n ≈ [nly n2y /z3]£ Z 3 there exist lineary independent vectors p, q £ Z3 such that pn =  qn =  Ö, andh (p) h (q )< ^ -| -1  (∏). if each of the numbers \nx |, | |, j /z3 1 is lessthan the sum of the two others ; 4x /г (p )^ (q )≤ ^ (n ) ,  otherwise.P r o o f .  We may assume without loss of generality that 0≤/z1 ≤ « 2 ≤ ^ 3 >0 . In virtue of Lemmata 1 and 5 the area A (K ) of the domainK:. l A ' I ≤ l ,  |K|≤1, \— X —- 2-Y \ ≤ \1 1  1 1  1 ∏s лзsatisfies(15)∕  24Л ( К ) > 4 /—---- 2----- 5 n3, if пх +  п%> п а,Q\ +  n2 +  n3A (K) =  4, otherwise.Let a, b be a basis, the existence of which is asserted in Lemma 3. The affine transformation X = a lx-^b1y y Y = a 2x-\-b2y  transforms the domain К into the domainK ':  \atx + b ty  |≤1 ( t ≈ h  2 , 3)satisfying(16) ∙ Л ( К ' ) = Л ( Ю  (” l∙ ”2' " з) ∙Л3In virtue of Minkowski's sccond theorem there exist two linearly independent integer vectors [x l9 y x] and [x2, y2] such that(17) I (XiXj 4- bt yj I ≤  Xj (/= 1, 2, 3; j — 1, 2) and(18) X1?ia/1 (K ')≤4.Putting p =  ax1~{-by1, q =  ax2 +  b y2, we infer that p, q are linearly independent, satisfy p∏^qn =  0  and in virtue of (15), (18)A ( p ) A ( q ) ≤ V * ;| < ^/-§-/(n), if n1-}-n2> n 3,( ≤ n 3, otherwise.P r o o f  o f  T h e o r e m  1. If n =  [slt e2, s3], where ez∙£ {l ,  -r-1}, it suffices to take p≈=[elf s2, OJ, q = [0 , 0, e3]. If n4 =[elf e2, e3] for every choice of elf e2, s3, then by Lemma 4 there exists a vector m ^ Z 3 satisfying the conditions(19) mn =  0,( 2 0 ) 0 </ (m )<v/ 2 A(n).Now, by Lemma 6  applied with n replaced by m there exist vectors p, q f Z 3 such that(2 1 ) pm =  qm =  0 , dim(p, q ) = 220 S. Chatadus, A. Schinzel∕and( 2 2 ) h (p) h (q )<  max  ^(m)» h (m)}.-The equations ( 2 0 ) and (2 2 ) imply that n =  //p4-^q ; и, while the inequa­lities (20) and (22) imply that h (p) h ( q ) <  [(4/3) h (n)]1/2.It follows that the number c0(3) defined in [5] by the formulak—2c0 (ß) =  sup inf h (p) h (q) h (∏ y -1n t Z k p, q(Z*∏4=0 dim (p, q ) =  2/ r= ap + vq , и, V £ Qsatisfies c0(3)≤,\J4/3 and if c0(3) =  \j4/3y the supremum occurring in the defi­nition of c0(&) is not attained. By Theorem 2 of [5] there exist vectors p0, q0 £Z3 linearly independent and such that n = « 0Po+-^oqo» vo£ Z, and Ä(p0)Ä(q0) <[(4/3)/г(п)]1'2. The proof *of Theorem 1 is complete.The proof of Theorem 2 is again based on several lemmata^ We shall set for t =  1 , 2, 3, . . .n, =  [(2t2 4- 2t)(fit2 4 - 4t— 1), (2<2 +  2t)(6t2 4 - 6 t -  1),(4*2 -j- 402-  (2^2- 1 )(2<2 4-2<— 1)],and for vectors m, p, . . .  we shall denote the v-th coordinate by mv, pv respec" tively.L e m m a  7. If∏/∏i =  0 , m£Z3, 0 < h (m )≤ 8 t2 +  8t—2y then we have m==m/ for an / ≤ 6 , where . ∕m i=  [o 2^ -f-6 * - l ,  —(6tf2 4-4<— 1 ), 0 ], m2 =  [2 /2 4 - 2 /— 1 , —(4^4-4Q, 2t2+ 2 t ]yme= [4 <2 +  4/, - (2 Г - - 1 ) ,  — {2t2-\-2t)]y ∏u =  [2<2 + 2^4-1, 2*8 4-4*+ l. -(4*?4-.4*)j,m 6 =  [2,6<24-8 < 4 -1, _ ( 6 * 2 +  6*)] (<4=1), m 0 =  [6f24 -6<4-1, 4^4-2, - ( 6 ^  + 601-P r o  of. The vectors Ш/( l ≤ / ^  6) all satisfy the equationnr∏i≈O. Since the vectors m1 and m2 are linearly independent, every vector m (Z 3 satisfying nm =  0  is of the form umx-\-vm2i и, v iQ .Let u =  ajc, v = b jc y a, b, с ( Z, (a, by c ) = 1 , c>0 . It follows from с \'ати +  Ьта1, c\am^j+bni2j  that c\(ay b){tnlim2i—m2im\j)y hence, c\mum2j —m2imlj  ( 1  ≤ i  < j ≤  3).But (mnm23 —m2lm13f ml2m23—m22m13)== m23(m lly m12)= m 23 and (m23y miu т22—т21У rnl2)≈ (m 23y m2V m12)=  1 , hence, c — 1 and we get m ^am ^ötr ia . Considering the third coordinate, we find ] b | (2t2 +  2 t )≤8 t2-\-8t — 2, hence, 1*1=3. .Considering the first coordinate, we get|а(6 <2 4 - 6 < - 1 ) 4- 0 (2 <?4 - 2 < ~ 1 ) | ≤ 8 *2 4 - 8 < - 2 ;I a I ( 6 / 2 +  6 *— 1)≤  8t2 4- 8 t - 2 4 - 1 b | (212 4- 2t- 1 )  ≤  14t2 4-141 - 15,hence, la |≤1 or a = ± 2 ,  b ≈  3. For a =  0 we get m ≈ 6  [2t2-\-2t— 1, — (4t2-\-4t), 2t2 4 - 2t] =  ±  m2. For | a [=  1 the inequality for the second coordinate ∙I а (612 4- 4t— 1)4-^ (412 4- 4t) | ≤  8t2 4- 8t -  2gives b≈ 0 or ab<0. For a = ± l ,  b =  0 we get m ≈ i ∏ i f ,  for a = ± l , & = = F l  we get m =  ± m 3; for a== ±  1 , b =  =f2  we get m = ± m 4; for a = ± l ,  £==F3 we get m = ± m 5; for a =  ±2 , b = + 3  we get m ≈ ± m 6.L e m m a  8 . If p, q f Z 3 are linearly independent and p m ^ q m ^ O ,  thenk(p)h(q)>4t2 +  4t.P r o o f .  p m ^ O  implies ^ ≈ O m o d  6 / 2 +  4/— l ,p a= 0  mod 6 /2 -f∙6 <— l.Hence∙ Р\=Ръ=®  or |/>2 l ≥ 6 ^ 2 +  6 /— J. Similarly, qt =  q2 =  0 or j q2|≥ 6 / 2 +  6 /— 1. Since p, q are linearly independent, A ( p ) A ( q ) ≥ 6 / 2 +  6^— l > 4 / 2 +  4/.L e m m a  9. If p, q£Z 3 are linearly independent andpm3 =  qm 2 =  0 ;then ,k (p )h (q )^ 4 t -  +  4t.P r o o f .  The equationр т а=(2£2+2/— \)px— (4/2+4/ )р 3 +  (2(г +  2 ( )р 3 =  0 ' gives p ^ O  mod 2t'2 +  2t— l, hence, /?x =  0 or \p\\>2t~+2t. The former possi­bility gives |p3 |≥2. Similarly, <7 i =  0, | q3 > 2  or |^x|≥2/ 2 +  2/. Since p, q are linearly independent, p1≈ q 1 =  0  is excluded, hence,h (p)A (q )≥m in  {2 (212 +  21), (21-+ 2ff}>4 t*  +  4/.L e m m a  10. If p, q ( Z 3 are linearly independent and pm3 ≈ q n i 3 =  0, thenh (p) h ( q ) ≥ 4 ^2 +  41.P r o o f .  The equationpm8 =  (4/2+ 4  t ) P l -  (2t2 - 1  )p 2-  (2 f -+ 2t) p3=0,gives p2≡≡t0 mod 2/ 2 +  2/, hence p2= 0 or \p2 | ≥ 2 / 2 +  2/. The further proof is similar to that of Lemma 9.L e m m a  11. If p£Z3, pm4 =  0, then either p =  0 or / r (p )>2/+ l.P r o o f .  The equationpm^ =  (2 t2-\-el t Jr 1) /?i +  (2/ 2 +  4/+ 1) p2—(4/ 2 +  At) /?3= 0gives(24)( (2 / 2 4- 2t)( р г+ р 2—2p3) + px +  (2/ +1  )p 2 =  0.If p l + p 2—2p3 =  0, then p l +  (2t +  l )p 2≈ 0  and either />i =  0 or |/>i|≥2/+l.If p\+p2—2/?3 ф0, then since by (24) p\≡≡p2 mod 2, we obtainp L+Pz -2p3=2s, s(:Z\ {0 },  /71 +  (2/+l)/?a= - ( 4 * 2 +4/)s.Hence, p3+ tp 2=  ~{2t-+2t-\-\)s andmax {| p21. 1 p31} ≥ 2 <2 ^ + 1> 2 .^thus / t (p )≥ 2 /4 -l.L e m m a  12. If p, q £ Z8are linearly independent and pm6 =  qm& =  0, then k(p)h(q)>4t2 +  4t (<4=1).P r o o f .  The equationpnig =  2/?! -I- (6t2 -h 8t 4- 1) p2—(&t2 +  6t) p3 =  0gives2pi 4- (214“ 1) Рч 4~ (612 4- 6>t)(p2—Рз) — 0.If p2= p s, we get Px≡≡O mod 2/4-1, hence, \px\ ≥ 2 i+ l .  If />аФ/>3 ’ we Set (2/4-3) max {I/?! |, \p2|}^:6/2 4-6/, hence,max {I />, I, |/72 |}≥-0 ^ > 3 / - 2A decomposition of integer vectors. II 2 1and A(p)≥:3/— 1. Similarly, q2 =  q3 and \qx \≥2t-\-1 or Ä (q )> 3 / — 1. Since p, q are linearly independent, p^—p3t Яч =  Яъ is excluded and we get for t =)=1h (p) h (q)≥min{(2/4- 1)(3/— 1), (3t— l ) 2}≥(2/4- 1)(3/- 1).L e m m a  13. If p, q ( Z 3 are linearly independent and pm6 =  qm6 =  0, thenh (p) h (q )≥4/24-4/.P r o o f .  The equationpm,; =  (6t2 4 - 6 /4 - l)/?iH~ (4∕  -f- 2 ) p2—(6/2 -f- 6/) p3 =  0gives(6/2 4- 6/)(/71 — p3) 4- P i  4- (4/ 4- 2) p2 =  0.If p x—/73=0, we get Pi≡≡O mod 4/4-2, hence, \px \≥4t+2∙К \Р\—Рг |≥2, we get(4/4-3) max {\p\\, \p21}≥2 (6/2-f-6/),hence,max {\px |, I pa > 3 1and ^ (p )≥3/4 - l.  If P\—  /73 =  zb 1 ∙ we get /?14-(4/4-2)/72==(6/24-6/), hence either|/»i|≥»4*+2 or ^a =  [q : - ^ ± | 0 - ]  or p a =  [ T  (6^ ^ -j- ] +  1.The last two formulae give the following possible values for T l A ,  ^ 2] :[3/, ~ \ ,  [ t - 1, i £ i ] ,  [ — ∕  — 2, [ - 3 / - 3 ,Hence, either h (p)≥3/4-2{//2} or /?i— /?3= ± 1  and p2=  [(3/4-2)/2]∙ Similarly, either h (q)≥3/'4-2{//2) or ^3= ± 1  and <72=  [(3/4-2)/2]. Since p, q are lin­early independent it follows thath (p) h (q)S>(3*+2‘{-\-})[Щ ?-)>№ +и.P r o o f  o f  T h e o r e m  2. Since4*2+4 ; / 42 2  S. Chatadus, A. Schinzellim v:v/ ( «2 +  4/)2- ( 2 ^ - l X 2 < 2+ 2 / - l )  V  3 for every e> 0  there exist t, such that(2 ) 4/a-b4/>^/(-|— e) h (n,)and we fix such a value of /.If n/≈//p4-^q, u, v t Q  and p, q ( Z 3 are linearly dependent, then since (nt\> nt2 » Л/з)=1* we have either p =  0 or p =  s∏r s £ Z \ {0 } ,  thus h (p )> h (n t)t and similarly for q. It follows that for рфО, q4=0h (p) h (q )≥ A  (n,)2 > V ( 4 ' - e) AIf p, q are linearly independent, then pxq4=0 and (p X q ) ∏/ =  0. On the other hand, either &(p) h (q )≥4/ 2 4-4/ or h (p X q )≤ 2 A -  (p) h (q) :< 2(4/24-4/— 14) ≈  8 /2 4 - 8 /—2. In the latter case in virtue of Lemma 7 we have p X q ≈ ∏ b  for na i ≤ 6 . Hence, /?m, =  =  0  and from Lemmata 8-13 we obtain h(p)h(q)≥:4t2+4t).In view of (25) the theorem follows.A decomposition of integer vectors. II 2 3Re ma r k .  There exist decompositions fy≈wp-t-i'q with A (p )A (q )  =  4/ 2 +  4^ , namely n, =  (6 /2 +4^— 1 )[2t2 +  2t, 0 , — (2 t2+2^ — 1)] +  (2t2+ 2t)(6t2+6^ — 1 ) [0 , 1 , 2 ] orn, =  (2*2 + 2 0 (6 *a+4<— 1 ) [1 , 0 , ' 2 ] +  (6 / 2 +  6 / - l ) [ 0 , 2 /2 -f-2 /, 1 - 2 *2].R E F E R E N C E S1. A. С h a t e 1 e t. Lemons sur la thcorie des nombres. Paris, 1913.2. A. S c h i n z e l .  Reducibility of lacunary polynomials. VII. Mh. Math., 102, 1986, 309-337,3. A. S c h i n z e l .  A  decomposition of integer vectors. /. Bull. Polish Acad. Sei. Math.,35, 1987, 155-159.Institute o f  Mathematics Received 14. 11. 1986Polish Academy ol Sciences Revised 13. 03, 1988ul. Sniadeckich 8 00-950 Warszawa, Poland

A decomposition of integer vectors. II.

http://sci-gems.math.bas.bg/jspui/bitstream/10525/2358/1/Pliska-11-1991-015-023.pdf

A decomposition of integer vectors. II.

Abstract

Similar works

Full text

Available Versions

Bulgarian Digital Mathematics Library at IMI-BAS