丢番图方程X~2-(a~2+1)Y~4=3-4a

设a≥2是正整数.本文证明了:当a=2时,方程X~2一(a~2+1)Y~4=3-4a仅有正整数解(X,Y)=(20,3);当a=3时,该方程仅有2组互素的正整数解(X,Y)=(1,1)和(79,5);当a≥4且4a+1非平方数时,该方程最多有4组互素的正整数解(X,Y);当a≥4且4a+1为平方数时,该方程最多有5组互素的正整数解(X,Y).     

丢番图方程X~2-(a~2+4p~(2n))Y~4=-4p~(2n)

令α,n≥1为整数,p为素数.本文证明了丢番图方程X~2-(a~2+4p~(2n))Y~4=-4p~(2n)以及X~2-(a~2+p~(2n))Y~4=-p~(2n)在一定条件下最多只有两组互素的正整数解(X,Y).     

关于方程ax+by=cz的Terai-Jesmanowicz猜想

设m是正整数,证明了:(A)如果b是奇素数,且a=m3-3m,b=3m2-1,c=m2+1,那么丢番图方程ax+by=cz(1)仅有正整数解(x,y,z)=(2,2,3);(B)如果b是奇素数,且a=m|m4-10m2+5|,b=5m4-10m2+1,c=m2+1,那么丢番图方程(1)仅有正整数解(x,y,z)=(2,2,5).     

关于方程ax+by=cz的Terai-Jesmanowicz猜想

设m是正整数,证明了(A)如果b是奇素数,且a=m3-3m,b=3m2-1,c=m2+1,那么丢番图方程ax+by=cz(1)仅有正整数解(x,y,z)=(2,2,3);(B)如果b是奇素数,且a=m|m4-10m2+5|,b=5m4-10m2+1,c=m2+1,那么丢番图方程(1)仅有正整数解(x,y,z)=(2,2,5).     

关于丢番图方程ax2+by2=cz2+dw2的整数解

数论中作为勾股定理的推广曾讨论过方程x2 +y2 =z2 +w2 ( 1 )的整数解 (如文 [1 ]、[2 ]) ,文 [2 ]得到了方程 ( 1 )满足 ( x,y,z,w) =1的全部整数解的一组公式 ,但表达式不够简洁。本文将其推广 ,考虑更一般的这类四元二次丢番都方程ax2 +by2 =cz2 +dw2 ( 2 )其中 a,b,c,d均为正整数 ,( a,b,c,d) =1。当知道它的一组不全为零的整数解时 ,来导出它满足 ( x,y,z,w) =1的全部整数解的公式。按所设 ,显然 z,w不会全为 0 ,不妨设 w≠ 0 ,从而方程 ( 2 )可变为a( xw) 2 +b( yw) 2 -c( zw) 2 =d令 X=x/ w,Y=y/ w,Z=z/ w,得a X2 +b Y2 -c Z2 =d …     

关于丢番图方程X~2-(a~2+1)Y~4=8-6a

设a是正整数.本文证明了:当a=1时,方程X~2-(a~2+1)Y~4=8~6a仅有正整数解(X,Y)=(2,1);当a=2时,该方程仅有正整数解(X,Y)=(1,1);当a=3时,该方程无正整数解(X,Y);当a=4时,该方程仅有2组互素的正整数解(X,Y)=(1,1)和(103,5);当a≥5且6a+1非平方数时,该方程最多有3组互素的正整数解(X,Y);当a≥5且6a+1为平方数时,该方程最多有4组互素的正整数解(X,Y).     

关于丢番图方程b~x+(2~α)~y=(b+2~α)~z

结合初等和高等的方法研究丢番图方程b~x+2~(αy)=(6+2~α)~z,α≥3正整数解的问题,并得到了如下结论:1.若b为平方数,则方程只有正整数解(x,y,z)=(1,1,1);若b+2~α为平方数,则x=1.2.若x1,则2■z.3.方程3~x+(2~(2k+1))~y=(3+2~(2k+1))~z,k■2 (mod 6),2k+1∈N,2k+11只有正整数解(x,y,z)=(1,1,1).     

Mordell方程y~3=x~2+2p~4的正整数解

设p是奇素数.对于非负整数r,设U_(2r+1)=(α~(2r+1)+β~(2r+1))/2~(1/2),V_(2r+1)=(α~(2r+1)-β~(2r+1))/6~(1/2),其中α=(1+3~(1/2))/2~(1/2),β=(1-3~(1/2))/2~(1/2).运用初等数论方法证明了:方程y~3=x~2+2p~4有适合gcd(x,y)=1的正整数解(x,y)的充要条件是p=U_(2m+1),其中m是正整数.当上述条件成立时,方程仅有正整数解(x,y)=(V(2m+1)(V_(2m+1)~2-6),V_(2m+1)~2+2)适合gcd(x,y)=1.由此可知:当p10000时,方程仅有正整数解(p,x,y)=(5,9,11),(19,1265,123),(71,68675,1683)和(3691,9677201305,4541163)适合gcd(x,y)=1.     

关于指数丢番图方程a~x+b~y=c~z的Terai猜想

本文证明了:当a=|m(m4-10m2+)|,b=5m4-10m2+1,c=m2+1,其 中m是偶数时,如果m≥542,则方程ax+by=cz仅有正整数解(x,y,z)=(2,2,5).     

关于丢番图方程x~2-D=p~n的解数

设D是正整数,p是适合p?D的奇素数。本文证明了:当max(D,p)≥10~(190)时,方程x~2-D=p~n至多有3组正整数解(x,n)。     

The equation $$x^4+2^ny^4=z^4$$ x 4 + 2 n y 4 = z 4 in algebraic number fields

Acta Mathematica Hungarica - Let n be a positive integer. We show that if the equation $$(1) \qquad \qquad \qquad x^4+2^ny^4=z^4$$ has a solution (x,y,z) in a cubic number field K with $$xyz \neq...     

Magische 4\times 4-Quadrate

Zusammenfassung. Diese Note behandelt zwei Typen von magischen -Quadraten und beschreibt mittels der Symmetriegruppen eines 4- bzw. 5-dimensionalen Würfels die Gesamtheit aller solchen Quadrate. Eingegangen 15.1.1997, angenommen 28.2.1997     

Locally 4-by-4 grid graphs

We investigate locally grid graphs. The main results are (i) a characterization of the Johnson graphs (and certain quotients of these) as locally grid graphs such that two points at distance 2 have precisely four common neighbors, and (ii) a complete determination of all graphs that are locally a 4 × 4 grid (it turns out that there are four such graphs, with 35, 40, 40, and 70 vertices).     

Le equazioni diofanteeax 4-bx 2+1=□, 4x 4+x 2 y 2+4y 4=□.

Summary Si determinano tre classi di quartiche u²=ax⁴+1, a e b interi, b²−4a≠□ per le quali l'appartenenza di un punto razionale (x₀,y₀),x₀≠0 porta l'esistenza nella stessa quartica di infiniti punti razionali. Si dimostra pure che l'equazione 4x²+x²y²+4y⁴=□ non ammette soluzioni intere con |xy| > 1. A BeniaminoSegre nel suo 70^mo compleanno. Entrata in Redazione il 26 febbraio 1973.     

Semi pandiagonal magic 4 × 4-squares

It is proved that the set of all semi pandiagonal 4 × 4-squares precisely consists of 3456 squares belonging to three orbits under the faithful action of a group of order 1152 which is a semi direct product of symmetric groups.     

Full 4-colorings of 4-regular maps

A full coloring of a planar map is a face coloring such that all the faces at ech vertex are colored differently. In this paper the planar 4-regular maps which have a full 4-coloring are characterized. This leads to a characterization of the planar maps (not necessarily 4-valent) which have a coupled 4-coloring.