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Open questions

OQ. 3890. 1). If a1, a2, ..., a9 ∈ {1,2, ...,9}and ai6=aj

(i, j ∈ {1,2,3,4,5,6,7,8,9}) then determine all polynomialsfk(n) =Z →Z (k= 1,2,3,4,5,6,7,8,9) such that

P9 k=1

afkk(n) is divisible bya1a2...a9 for all n∈N.

We have the following 22n+1+ 32n+1 0 (mod 5),2n+2+ 32n+1 0 (mod 7), 28n+3+ 3n+1 0 (mod 11),24n+2+ 3n+20 (mod 13),

26n+2+ 34n+2 0 (mod 17),23n+4+ 32n+1 0 (mod 19), 25n+1+ 3n+3 0 (mod 29),24n+1+ 36n+90 (mod 31)

2). Determine all polynomials fk(n) =Z →Z (k= 1,2,3,4,5,6,7,8,9) for which

P9 k=1

afkk(n) is divisible by 45 for alln∈N.

Mih´aly Bencze OQ. 3891. 1). Solve inQ the equation

(x+ 1) (y+ 1) (z+ 1) = (x1) (y1) (z1).If x= a+ba+b, y= b+cbc, z= c+aca, when a, b, c∈Q, then we obtain infinitely many solutions of the given

equation

2). Solve inZ the given equation 3). Solve inQ the equation

Qn k=1

(xk+ 1) = Qn k=1

(xk1) 4). Solve inZ the previous equation

Mih´aly Bencze OQ. 3892. Ifxk>0 (k= 1,2, ..., n) such that

Qn k=1

xk= 1,then determine all a, b, c >0 for which

Pn k=1

1

(a+xk)b + 1

1+

Pn k=1

xk

c (a+1)n b +(n+1)1 c.If a= 1, b= 2,c= 1, then we obtain a solution forn= 3.

Mih´aly Bencze OQ. 3893. 1). Solve inQ the equationxy+yz+zx=1. Ifx= 1a−bab, y= 1bbcc, z= 1ccaa,where a, b, c∈Q, then we obtain an infinitely many solutions for the given equation.

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3). Solve inQ the equation P

cyclic

x1x2...xn1 =1 4). Solve the previous equation inZ

Mih´aly Bencze OQ. 3894. Ifxk>0 (k= 1,2, ..., n) then

P

cyclic

xn11+x2...xn

xn21+x3...x1

n2n41

Pn

k=1

xk 2n−2

.

Mih´aly Bencze OQ. 3895. Ifxk>0 (k= 1,2, ..., n) then

P

cyclic

x1+x1

2 x2+x1

3

...

xn+x1

1

nn23

Pn

k=1

xk+ Pn k=1

1 xk

n2

.

Mih´aly Bencze OQ. 3896. Ifxk>0 (k= 1,2, ..., n) then

Pn k=1

xnk+n2 Qn k=1

xk n(n+1)

nn−12 ·

P

cyclic

x1x2

!n+1

2

Pn k=1

xk

.

Mih´aly Bencze OQ. 3897. Ifxk>0 (k= 1,2, ..., n) such that

Qn k=1

xk= 1,then determine all λ∈R for which

Qn k=1

1 +xλk

n(n2n1)λ

Pn

k=1

xk

(n1)λ

.If n= 3 then λ= 2 is a solution.

Mih´aly Bencze OQ. 3898. Ifxk >0 (k= 1,2, ..., n) such that

Pn k=1

xk =nthen determine all a, b, λ >0 for whicha

Pn k=1

xλk+b Qn k=1

xk≥na+b.The casea=n−1, b=n, λ= 2 is a solution.

Mih´aly Bencze

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OQ. 3899. Solve inZ the equation x2+yxy2+tz2 +y2+zyz2+tx2 +z2+xzx2+ty2 = t+23 . Mih´aly Bencze OQ. 3900. Solve in Z the equation 3 x2+y2+z2

=t2+u2+v2+w2.If t=x+y+z, u=x−y, v=y−z, w=z−x, wherex, y, z ∈Z, then the equation have infinitely many solutions inZ.

Mih´aly Bencze OQ. 3901. Solve in Z the equation

x31y1+x32y2+x33y3= 2 z21+z22+z32

+u1t21+u2t22+u3t23.If y1=x2+x3, y2 =x3+x1, y3 =x1+x2, z1 =x1x2, z2=x2x3, z3 =x3x1, u1 =x1x2, u2 =x2x3, u3=x3x1, t1 =x1−x2, t2 =x2−x3, t3=x3−x1, where x1, x2, x3∈Z then the given equation have infinitely many solutions inZ.

Mih´aly Bencze OQ. 3902. Solve in Z the equation

x51y1+x52y2+x53y3=z14t1+z42t2+z34t3+u21v1+u22v2+u23v3.

Ify1 =x2+x3, u2=x3+x1, y3 =x1+x2, z1=x1, z2=x2, z3=x3, t1 =x22+x23, t2=x23+x21, t3 =x21+x22, u1=x1−x2, u2=x2−x3, u3 =x3−x1, v1 =x31x2+x21x22+x1x32, v2=x32x3+x22x23+x2x33,

v3 =x33x1+x23x21+x3x21, wherex1, x2, x3∈Z,then the given equation have infinitely many solutions in Z.

Mih´aly Bencze OQ. 3903. Solve in Z the equation

2 x61+x62+x63

=y15z1+y52z2+y35z3+u21v1+u22v2+u23v3.

Ify1 =x1, y2=x2, y3 =x3, z1 =x2+x3, z2=x3+x1, z3 =x3+x2,

u1 =x1−x2, u2 =x2−x3, u3=x3−x1, v1=x41+x31x2+x21x22+x1x32+x42, v2 =x42+x32x3+x22x23+x2x33+x43, v3=x43+x33x1+x23x21+x3x31+x41 where x1, x2, x3∈Z, then the given equation have infinitely many solutions in Z.

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OQ. 3904. Solve in Z the equation 2

xk+21 +xk+22 +xk+23

+ 2 yk1z1+yk2z2+yk3z3

=

= 2

tk+11 u1+tk+12 u2+tk+13 u3

+v21w1+v22w2+v32w3.

Ify1 =x1, y2=x2, y3 =x3, z1 =x2+x3, z2=x3+x1, z3 =x2+x3, t1 =x1, t2 =x2, t3 =x3, u1 =x2x3, u2 =x3x1, u3=x1x2, v1 =x1−x2, v2 =x2−x3, v3=x3−x1, w1 =−xk1 +xk2 +xk3, w2=−xk2+xk3+xk1, w3=−xk3+xk1+xk2,where x1, x2, x3∈Z, then the given equation have infinitely many solutions in Z.

Mih´aly Bencze OQ. 3905. Solve in Z the equationn

Pn k=1

x2k =y2+ P

1i<jn

zij2.If y=

Pn k=1

xk, zij =xi−xj, i, j∈ {1,2, ..., n} when xi ∈Z (i= 1,2, ..., n),then the given equation have infinitely many solutions inZ.

Mih´aly Bencze OQ. 3906. 1). Solve inQ the equation

Pn k=1

1

x2k = Pnn2 k=1

y2k

+n1 P

1i<jn

zij2.If yk=xk (k= 1,2, ..., n), zij = xxi

j xxji, where xi ∈Q (i= 1,2, ..., n) then the given equation have infinitely many solution inQ

2). Solve inZ the given equation

Mih´aly Bencze OQ. 3907. Solve in Z the equation 2

Pn

k=1

x2k−y1y2−...−yny1

= Pn k=1

zk2. Ifyk =xk, zk=xk−xk+1,where xk∈Z (k= 1,2, ..., n), then the equation have infinitely many solutions inZ.

Mih´aly Bencze

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OQ. 3908. Solve in Z the equation

x21+x22+...+x2k x1+x2+...+xk +x

2

2+x23+...+x2k+1

x2+x3+...+xk+1 +...+x

2n+x21+...+x2k1 xn+x1+...+xk1 =

n Pn i=1

x2i Pn i=1

xi

,when k∈ {1,2, ..., n}.The casex1 =x2=...=xn∈Z offer infinitely many solutions.

Mih´aly Bencze OQ. 3909. Solve in Z the equation

(n1)xn11x2x3...xn

xn11(x1x2)n21+xn21 +...+ (n1)xn−1n x1x2...xn−1

xnn1(xnx1)n21+xn11 =n.

The casex1=x2 =...=xn∈N offer infinitely many solutions in Z.

Mih´aly Bencze OQ. 3910. Ifxk>0 (k= 1,2, ..., n) such that

Qn k=1

xk= 1 then determine all a(n), b(n)>0 for which

Pn k=1

x2k−n≥a(n) Pn k=1

xk−b(n) P

1i<jn

xixj.If n= 3, then a(n) =b(n) = 18.

Mih´aly Bencze OQ. 3911. 1). Solve inQ the following system:







 Pn k=1

xk=n (n1)

Pn k=1

x2k+n Qn k=1

xk=n2 2). Solve inZ the given system.

Mih´aly Bencze

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OQ. 3912. Solve in Qthe following equation Qn

k=1

xk+ n sQn

k=1

x+ykn

= n(n2(x+2)1) P

1≤i<j≤nzizj.

Mih´aly Bencze OQ. 3913. Determine all n, k∈N for which nk+ 1 = aa...a| {z }

mtime

,where a∈ {1,2,3,4,5,6,7,8,9}.An example is 65+ 1 = 7777.

Mih´aly Bencze OQ. 3914. Determine all n, k∈N for which 1k+1k

2k+1k

... nk+k1 is the square of a rational number.

Mih´aly Bencze OQ. 3915. Determine all n, k∈N for which

Φ Φ nk

+ Φ (Φ (kn)) =n+k,where Φ is the Euler‘s totient function.

Mih´aly Bencze OQ. 3916. Ifxk(0, π) (k= 1,2, ..., n) and

Pn k=1

xk=π then determine all functions f : (0, π)(0,+) such that

Pn k=1

f(xk) = Pn

k=1 1 f(xk)

Qn

k=1

f(xk).

A solution isf(x) =tgx.

Mih´aly Bencze OQ. 3917. Letf : [a, b]→R be a convex function. Determine all λ∈[0,1]

such that b1a Rb a

f(x) λ(f(a)+f(b))

2 + (1−λ)f a+b2

. A solution is λ= 12. Mih´aly Bencze OQ. 3918. Let be P(x) =a0xn+a1xn1+...+an1x+an.Determine all a0, a1, ..., an∈C for which all positive rational number AB (A, B∈N) can be represented in following form: BA =

Qn i=1

(P(i))±1.ForP(x) = x(x+1)2 we have a solution.

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OQ. 3919. All positive integers mcan be represented in the following forms:

1). m=±dk(1)±dk(2)±...±dk(n) 2). m=±σk(1)±σk(2)±...±σk(n) 3). m=±Φk(1)±Φk(2)±...±Φk(n) 4). m=±Ψk(1)±Ψk(2)±...±Ψk(n) 5). m=±F1k±F2k±...±Fnk

6). m=±Lk1 ±Lk2 ±...±Lkn 7). m=± n1k

± n2k

±...± nnk

Mih´aly Bencze OQ. 3920. Determine all a1, r∈C,ak=a1+ (k1)r, and allp∈N for which all positive integers m can be represented in the following form:

m=±ak1±ak2±...±akn.

Mih´aly Bencze OQ. 3921. Determine all b1, q∈C, bk=b1qk1 and allp for which all positive integersm can be represented in the following form:

m=±bk1±bk2±...±bkn.

Mih´aly Bencze OQ. 3922. Determine all ak>0 (k= 1,2, ..., n) for which P

cyclic a1a2

a2+a3 0.

Mih´aly Bencze OQ. 3923. Determine all ak>0 (k= 1,2, ..., n) for which

1 +

Qn k=1

ak

P

cyclic 1 a1+a2a3...an

!

≥n.

Mih´aly Bencze OQ. 3924. Determine all ak>0 (k= 1,2, ..., n) such that

Pn k=1

1

ak + 9P 1

a1+a2+a3 + 25P 1

a1+a2+a3+a4+a5 +...≥

4P 1

a1+a2 + 16P 1

a1+a2+a3+a4 +...

Mih´aly Bencze

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OQ. 3925. Ifxk>0 (k= 1,2, ..., n),then Pn

k=1

xnk+n Qn k=1

xk n21 P

cyclic

x1...xn1(x1+...+xn1).

Mih´aly Bencze OQ. 3926. Ifxk>0 (k= 1,2, ..., n),then

P

cyclic

x1+x

n1 2

xn32

n1

n·2

n−1 Pn k=1

x2nk 3 Pn

k=1

xn−2k

.

Mih´aly Bencze OQ. 3927. Ifxk>0 (k= 1,2, ..., n),then determineF(n)>0 such that

Pn k=1

xnk−n Qn k=1

xk ≥F(n) (x1−x2) (x2−x3)...(xn−x1).We haveF(1) = 0, F(3) = 4.

Mih´aly Bencze OQ. 3928. Ifxk>0 (k= 1,2, ..., n),then P

cyclic

xn+11

xn1+...+xnn1 n11 Pn

k=1

xk. Mih´aly Bencze OQ. 3929. Ifxk>0 (k= 1,2, ..., n),then

Pn

k=1

xk Pn k=1

1 xk

≥n+ 2 +

(n2)(n+1) Pn k=1

xnk1 P

cyclic

x1...xn−1 .

Mih´aly Bencze OQ. 3930. Ifxk>0 (k= 1,2, ..., n),then P

cyclic

(x2+...+xn)n1

xn11+x2...xn n(n21)n1. Mih´aly Bencze OQ. 3931. Ifxk >0 (k= 1,2, ..., n),then

nn−2 Pn k=1

x(n−1)2k Pn

k=1

xn−1k

n1 +

P

cyclic

x1...xn1

Pn k=1

xnk1 2.

Mih´aly Bencze

(9)

OQ. 3932. Ifxk>0 (k= 1,2, ..., n),then P

cyclic

x2+...+xn

(n1)xn−11 +x2x3...xn Pn(nn 1)

k=1

xnk2

.

Mih´aly Bencze OQ. 3933. Ifxk>0 (k= 1,2, ..., n),then

P

cyclic 1

xn−11 +x2...xn

(n1) Pn

k=1

xk P

cyclic 1 x1+...+xn−1

!

2 P

cyclic

x1...xn1 .

Mih´aly Bencze OQ. 3935. Ifxk>0 (k= 1,2, ..., n),then

n(n1) P

cyclic x1

x2+...+xn Pn

k=1

xk Pn k=1

1 xk

.

Mih´aly Bencze OQ. 3936. Ifxk>0 (k= 1,2, ..., n),then P

cyclic x21 x2 n+1

s nn

Pn

k=1

xn+1k

.

Mih´aly Bencze OQ. 3937. Ifxk>0 (k= 1,2, ..., n),then

Pn k=1

x2(n−1)k P

cyclic

x1x2...xn1 +

n Qn k=1

xk

Pn k=1

xk

2n Pn

k=1

xnk1.

Mih´aly Bencze OQ. 3938. Ifxk>0 (k= 1,2, ..., n),then determine alla, b >0 such that

a Pn k=1

xnk Pn k=1

xnk1

+

b P

cyclic

(x1x2)n1 Pn

k=1

x2nk 3 a+bn Pn

k=1

xk.

Mih´aly Bencze OQ. 3939. Ifxk>0 (k= 1,2, ..., n),then

Pn k=1

xnk Qn k=1

xk

+

2nn Qn k=1

xk Pn

k=1

xk

n ≥n+ 2.

(10)

OQ. 3940. Ifxk>0 (k= 1,2, ..., n),then

Pn

k=1

xk Qn

k=1

xk Pn

k=1

xn+1k

+

(n+1)nn1 Pn k=1

x

n(n1) 2

Pn k

k=1

xk Pn

k=1

x2k

...

Pn

k=1

xnk1

≥n+ 2.

Mih´aly Bencze OQ. 3941.P Ifxk>0 (k= 1,2, ..., n),then

cyclic

1

(x1+x2+...+xn1)n1 (n1)n1 Pn2

cyclic

x1x2...xn1.

Mih´aly Bencze OQ. 3942. Ifak>0 (k= 1,2, ..., n),then

P a1

a2+...+an +

Qn k=1

ak

(n1) Pn k=1

ank n(n−1)n2+1 .

Mih´aly Bencze OQ. 3943. Ifxk>0 (k= 1,2, ..., n),then P

cyclic

x21+...x2n1 x1+...+xn−1 n

Pn k=1

x2k Pn k=1

xk

.

Mih´aly Bencze OQ. 3944. Ifxk>0 (k= 1,2, ..., n),then

Pn k=1

xnk1 P

cyclic

x1x2...xn1 +

(n−1)n Qn

k=1

xk

Q

cyclic

(x1+...+xn1) 2.

Mih´aly Bencze OQ. 3945. Ifpk (k= 1,2, ..., n) are given prime, then solve inN the

equation Qn4n k=1

pk

=

3n

P

k=1 1 xk.

Mih´aly Bencze

(11)

OQ. 3946. If 0< xk< π, pk >0,(k= 1,2, ..., n), An=

Pn k=1

pkxk

Pn k=1

pk

,

G= Qn

k=1

xpkk Pn1

k=1pk

,then sinGGPn

k=1

pk

Qn

k=1

sinxk

xk

pk

sinAAPn

k=1

pk

. Mih´aly Bencze OQ. 3947. Solve in Qthe equation 2z4= x3+y3

x3+ 4y3 .

Mih´aly Bencze OQ. 3948. Solve in Qthe equationx3+nx+n+ 1 = (n+ 1)y2 when n∈N is given.

Mih´aly Bencze OQ. 3949. Solve in Z the equationy6 = 2x41

3x41

5x41 . Mih´aly Bencze OQ. 3950. Solve in Z the equationa

Pn k=1

x2k+b P

1i<jn

xixj =c when a, b, c∈Z are given.

Mih´aly Bencze OQ. 3951. Ifp is a given prime, then solve inZ the equationx2+p=yn.

Mih´aly Bencze OQ. 3952. Solve in Z the equation

x42y4

z48t4

u33uv2−v3

= 1.

Mih´aly Bencze OQ. 3953. Solve in Z the equation x3+x2y+axy2+b=cwhen a, b, c∈Z are given.

Mih´aly Bencze

(12)

OQ. 3954. Determine all n∈Z for which the equation x4+y4+nz4 = 1 have infinitely many solutions inQ.

Mih´aly Bencze OQ. 3955. Solve in Qthe equation

Pn k=1

x2pk =pwhen p∈Z is given.

Mih´aly Bencze OQ. 3956. Ifa, b∈Z are given, then solve inZ the equation

Qn k=1

x2k+a2

= ay2+b2n

.

Mih´aly Bencze OQ. 3957. Solve in Qthe equation

Pn k=1

x3k =p,when p is a given prime.

Mih´aly Bencze OQ. 3958. Ifa, b, c, ak, bk, ck∈Z (k= 1,2, ..., n) are given, then solve the equation

Qn k=1

akx2+bky2+ckz2

= ax2+by2+cz2n

.

Mih´aly Bencze OQ. 3959. Ifak∈Z (k= 1,2, ..., n) are given, then solve in Q the equation

Pn k=1

akx2p+1k = 0.

Mih´aly Bencze OQ. 3960. Solve in Z the equation

Pn k=1

x2pk = Pn k=1

yk2p whenp∈Z.

Mih´aly Bencze OQ. 3961. Ifak, bk, ck, dk ∈Z (k= 1,2, ..., n) are given, then solve in Qthe equation xy

Qn k=1

akx2+bky2

=zt Qn k=1

ckz2+dkt2 .

Mih´aly Bencze

(13)

OQ. 3962. Ifpk∈Z (k= 1,2, ..., n) are given prime, then solve inZ the equation

Pn k=1

pkx2k=y3.

Mih´aly Bencze OQ. 3963. Solve in Z the equations

1). x2n+y2n=zn 2). x2n−y2n=zn

Mih´aly Bencze OQ. 3964. Solve in Z the equationx2n+y2n=nzn.

Mih´aly Bencze OQ. 3965. Solve in Z the equationx2n 2nn

xnyn+y2n=zn.

Mih´aly Bencze OQ. 3966. Solve in Z the equationa bx2+cy22

+z2= dt2+eu22

wherea, b, c, d, e∈Z are given.

Mih´aly Bencze OQ. 3967. Ifb, ak ∈Z (k= 1,2, ..., n) are given, then solve in Z the

equation Pn k=1

akx2k =b Qn k=1

xk.

Mih´aly Bencze OQ. 3966. Solve in Z the equation

Pn k=0

ak nk

xnkyk=zn whereak ∈Z (k= 1,2, ..., n) are given.

Mih´aly Bencze OQ. 3969. Ifp, q are prime, andk∈Z is given, then solve inZ the

equation x2+py2 =k z2+qt2 .

Mih´aly Bencze

(14)

OQ. 3970. LetABC (a6=b6=c) be a triangle. Compute St(a, b, c) = minPat+bt

at−bt,when t∈R.We have S1(a, b, c) = 5, S2(a, b, c) = 3.

Mih´aly Bencze OQ. 3971. Compute Sk= P

1i<jn 1

(ji)(ji)k.

Mih´aly Bencze OQ. 3972. IfSnp = P

k0 n pk

then determine a relation betweenSn1, Sn2, ..., Snp.

Mih´aly Bencze OQ. 3973. IfRpn= P

k0 1

(pkn) then determine a relation between Rn1, R2n, ..., Rpn.

Mih´aly Bencze OQ. 3974. Ifak>0 (k= 1,2, ..., n) andλ≥t >0 then determine all tsuch that

Pn k=1

aλ+1k Pn k=1

aλk

Pn k=1

aλ+1k Pn k=1

aλ+1k t

1 t

. (For t= 2 we have a solution).

Mih´aly Bencze OQ. 3975. Denotean the nearest integer of k

n (k2).Compute 1).

Pn i=1

ai 2).

Pn i=1

1 ai

3). a 1

1a2...ai +a 1

2a3...ai+1 +...+a 1

na1...ai1 wherei∈ {2,3, ..., n1}.

Mih´aly Bencze OQ. 3976. Ifxk>0 (k= 1,2, ..., n),then determine alln∈N for which

P

cyclic

xn11−x2x3...xn

xn11+ (n1)x2x3...xn

n11

0.

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One of the most widely used solution is to solve the problem in the real space and they apply additional iteration steps (so-called cutting-plane algorithms or Gomory’s cuts)

Theorem 7 For a given complete graph K n , all connected labeled chordal graphs, which are equivalent to all edge subsets of K n inducing connected chordal graphs, can be enumerated

Remember that we have learnt a polynomial time algorithm for finding maximum matching in a bipartite graph?. The formalization works for all kind of graphs so it looks like a

A 2D similarity transformation model is used to project model features into the image as in [16] , though it does not describe the transformation between two per- spective