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Arithmetic properties of polynomials
Periodica Mathematica Hungarica ( IF 0.8 ) Pub Date : 2020-03-17 , DOI: 10.1007/s10998-020-00333-2
Yong Zhang , Zhongyan Shen

First, we prove that the Diophantine system $$\begin{aligned} f(z)=f(x)+f(y)=f(u)-f(v)=f(p)f(q) \end{aligned}$$ f ( z ) = f ( x ) + f ( y ) = f ( u ) - f ( v ) = f ( p ) f ( q ) has infinitely many integer solutions for $$f(X)=X(X+a)$$ f ( X ) = X ( X + a ) with nonzero integers $$a\equiv 0,1,4\pmod {5}$$ a ≡ 0 , 1 , 4 ( mod 5 ) . Second, we show that the above Diophantine system has an integer parametric solution for $$f(X)=X(X+a)$$ f ( X ) = X ( X + a ) with nonzero integers a , if there are integers m , n , k such that $$\begin{aligned} {\left\{ \begin{array}{ll} (n^2-m^2)(4mnk(k+a+1)+a(m^2+2mn-n^2)) &{}\equiv 0\pmod {(m^2+n^2)^2},\\ (m^2+2mn-n^2)((m^2-2mn-n^2)k(k+a+1)-2amn) &{}\equiv 0\pmod {(m^2+n^2)^2}, \end{array}\right. } \end{aligned}$$ ( n 2 - m 2 ) ( 4 m n k ( k + a + 1 ) + a ( m 2 + 2 m n - n 2 ) ) ≡ 0 ( mod ( m 2 + n 2 ) 2 ) , ( m 2 + 2 m n - n 2 ) ( ( m 2 - 2 m n - n 2 ) k ( k + a + 1 ) - 2 a m n ) ≡ 0 ( mod ( m 2 + n 2 ) 2 ) , where $$k\equiv 0\pmod {4}$$ k ≡ 0 ( mod 4 ) when a is even, and $$k\equiv 2\pmod {4}$$ k ≡ 2 ( mod 4 ) when a is odd. Third, we get that the Diophantine system $$\begin{aligned} f(z)=f(x)+f(y)=f(u)-f(v)=f(p)f(q)=\frac{f(r)}{f(s)} \end{aligned}$$ f ( z ) = f ( x ) + f ( y ) = f ( u ) - f ( v ) = f ( p ) f ( q ) = f ( r ) f ( s ) has a five-parameter rational solution for $$f(X)=X(X+a)$$ f ( X ) = X ( X + a ) with nonzero rational number a and infinitely many nontrivial rational parametric solutions for $$f(X)=X(X+a)(X+b)$$ f ( X ) = X ( X + a ) ( X + b ) with nonzero integers a , b and $$a\ne b$$ a ≠ b . Finally, we raise some related questions.

中文翻译:

多项式的算术性质

首先,我们证明丢番图系统 $$\begin{aligned} f(z)=f(x)+f(y)=f(u)-f(v)=f(p)f(q) \end {aligned}$$ f ( z ) = f ( x ) + f ( y ) = f ( u ) - f ( v ) = f ( p ) f ( q ) 对于 $$f(X) 有无穷多个整数解=X(X+a)$$ f ( X ) = X ( X + a ) 非零整数 $$a\equiv 0,1,4\pmod {5}$$ a ≡ 0 , 1 , 4 ( mod 5 )。其次,我们证明上面的丢番图系统有一个整数参数解 $$f(X)=X(X+a)$$f ( X ) = X ( X + a ) 和非零整数 a ,如果有整数m , n , k 使得 $$\begin{aligned} {\left\{ \begin{array}{ll} (n^2-m^2)(4mnk(k+a+1)+a(m^ 2+2mn-n^2)) &{}\equiv 0\pmod {(m^2+n^2)^2},\\ (m^2+2mn-n^2)((m^2- 2mn-n^2)k(k+a+1)-2amn) &{}\equiv 0\pmod {(m^2+n^2)^2}, \end{array}\right。} \end{aligned}$$ ( n 2 - m 2 ) ( 4 mnk ( k + a + 1 ) + a ( m 2 + 2 mn - n 2 ) ) ≡ 0 ( mod ( m 2 + n 2 ) 2 ) , ( m 2 + 2 mn - n 2 ) ( ( m 2 - 2 mn - n 2 ) k ( k + a + 1 ) - 2 amn ) ≡ 0 ( mod ( m 2 + n 2 ) 2 ) ,其中$$当 a 为偶数时 k\equiv 0\pmod {4}$$ k ≡ 0 ( mod 4 ),当 a 为奇数时 $$k\equiv 2\pmod {4}$$ k ≡ 2 ( mod 4 )。第三,我们得到丢番图系统 $$\begin{aligned} f(z)=f(x)+f(y)=f(u)-f(v)=f(p)f(q)=\ frac{f(r)}{f(s)} \end{aligned}$$ f ( z ) = f ( x ) + f ( y ) = f ( u ) - f ( v ) = f ( p ) f ( q ) = f ( r ) f ( s ) 对于 $$f(X)=X(X+a)$$ f ( X ) = X ( X + a ) 具有非零有理数的五参数有理解a 和无穷多个非平凡有理参数解 $$f(X)=X(X+a)(X+b)$$ f ( X ) = X ( X + a ) ( X + b ) 具有非零整数 a , b 和 $$a\ne b$$ a ≠ b 。最后,我们提出一些相关的问题。
更新日期:2020-03-17
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