关于Diophantine 方程xn+2kyn=pz2

设p 为奇素数且对任意的整数m, d, p≠(2^m±1)=/d², 则对任意的素数n > p^8p2, 方程xⁿ+2^kyⁿ=pz², k≥2 没有整数解(x, y, z) 使得x, y, z 两两互素且均不为0.     

Extensions of the D(?k2)-triples {k2, k2 ± 1, 4k2 ± 1}

Let n be a nonzero integer. A set of m distinct positive integers is called a D(n)-m-tuple if the product of any two of them increased by n is a perfect square. Let k be a positive integer. In this paper, we show that if {k ², k ² + 1, 4k ² + 1, d} is a D(−k ²)-quadruple, then d = 1, and that if {k ² − 1, k ², 4k ² − 1, d} is a D(k ²)-quadruple, then d = 8k ²(2k ² − 1).     

Y2=X3-p3上的无穷阶点

首先用(12)上的模函数参数化椭圆曲线y²=x³+1,然后利用虚二次域Q((-p)^1/2)上的类域论和Shimura互反律构造了当p≡7(mod24)为一素数时Y²=X³-p³上的一个无穷阶点.     

Small solutions of the congruencea1x12 +a2x22 +a3x32 +a4x42 ≡c(modp)

For any integersa ₁,a ₂,a ₃,a ₄ andc witha ₁ a ₂ a ₃ a ₄≢0(modp), this paper shows that there exists a solutionX=(x ₁,x ₂,x ₃,x ₄) ∈Z ⁴ of the congruencea ₁ x ₁ ² +a ₂ x ₂ ² +a ₃ x ₃ ² +a ₄ x ₄ ² ≡c(modp) such that

关于Pell方程组x2-(m2-1)y2=z2-(n2-1)y2=1

设m,n,L为正整数,本文证明了:如果mε,ε∈(0,1),且m>(123789LL(1/2))⁽1/(1-ε)),或j>10.25×10¹²log⁴(2(L+1)(123789LL(1/2))⁽1/(1-ε))),Pell方程组x²-(m2-1)y²=z²-(n²-1)y²=1的正整数解满足1≤k≤δL²,这里δ∈[1/2(123787LL(1/2))⁽1/(ε-1)),1],以及■且j=k=1或k+2≤j<1/3(5-2ε)k,2|(j+k),k>3/(1-ε),并改进了文[Proc.Amer.Math.Soc.,2015,143(11):4685-4693]的结果.     

探究x0x+y0y=r2与x0x+y0y=x02+y02的几何意义

问题背景苏教版教材必修二P105有这样一道习题:已知圆C的方程是x2+y2=r2,求经过圆C上一点M(x0,y0)的切线的方程.同学们在处理该问题时给出以下的解答过程:如图1,设切线的斜率为k,半径OM的斜率为k1,因为圆的切线垂直于过切点的半径,于是k=     

Carmichael numbers with a totient of the form a2?+?nb2

Let φ be the Euler function. Fix , and let be an arbitrary set of primes of positive lower natural density. Using a variant of the Alford–Granville–Pomerance construction, we show that there are infinitely many Carmichael numbers N with a totient of the form , where and is a nonempty product of primes from the set . In particular, for any fixed natural number n, there are infinitely many Carmichael numbers N such that φ(N) = a ² + nb ² for some positive integers a and b.