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LAST INTEGER SOLUTION IS THE BEST FOUND! 以上整数解是最优解!
For problems with integer constraints, you need to decrease the Tolerance setting in the Solver Options dialog box so that Solver can find a better integer solution. 对于具有整数约束条件的问题,应该减小“规划求解选项”对话框中的“允许误差”的设置,使“规划求解”找到更好的整数解。
Let r be an odd integer with r>1.In this paper the author gives a necessary condition for(X,Y,Z) being a positive integer solution of the equation X~2+Y~2=Z~r with Y being a power of an odd prime. 设r是大于1的奇数;给出了方程X2+Y2=Zr的正整数解(X;Y;Z)中Y为奇素数方幂的必要条件.
Let(a,b,c) be a primitive Pythagorean triple with a is even. In this paper we prove that if c is a prime power,then the equation x~2+b~y=c~z has only the positive integer solution(x,y,z)=(a,2,2) with y is even. 设(a;b;c)是一组适合a为偶数的本原商高数;该文证明了:当c是素数方幂时;方程x2+by=cz仅有正整数解(x;y;z)=(a;2;2)可使y是偶数.
In this paper, let p be an odd prime with p>3, we prove that the equation (xp-yp)/(x-y)=z2 has only the positive integer solution (x, y, z, p)=(3,1,11,5), satisfying x>y+1. gcd (x, y)=1. As a result, x is an odd prime power. 设p是大于3的奇素数,证明:方程2)()(zyxyxpp=--,1+>yx,1),gcd(=yx仅当p=5时有正整数解)11,1,3(),,(=zyx可使x是奇素数的方幂。
Using the elementary methods, all the positive integer solutions of the equation are obtained.The solvability of the equation is solved completely. 利用初等方法，获得了方程的所有正整数解，完全解决了该方程的可解性问题。
And based on these results the expressions for some binary indefinite quadratic equations integer solutions are obtained. 并且利用其结果得到了几类二元二次不定方程的整数解表达式。
number of positive integer solution 正整数解数
In this paper we have found out all integer solutions of the Diophanine equation in the title, which was proposed by Mordell in 1969 and reproposed by Guy in 1981. 1969年,Mordell提出求解不定方程6y~2=(x+1)(x~2-x+6),本文解决了这个问题,求得了这个方程的全部整数解。
In this paper,the author has proved that the Diophantine y(y+1)(y+2)(y+3)=nx(x+1)(x+2)(x+3) has no integer solutions when n=13~(2k),k is a natural number . 本文用初等方法证明了不定方程y(y+1)(y+2)(y+3)=nx(x+1)(x+2)(x+3)在n=13~(2k)(k为自然数)时无解.
It is proved that if p>3 and D is not divisible by p or primes of the form 2kp+1,then the equation x~p-2~p=pDy~2 has no positive integer solutions (x,y) with gcd (x,y)=1. 本文证明了:当p>3时;如果D不能被p或2kp+1之形素数整除;则方程xp-2p=pDy2没有适合gcd(x;y)=1的正整数解(x;y).
In 1997,Andrew Beal conjectured that if A=B=C=1,m,n,r are all larger than 2,the equation has not positive integers solution. 1997年;Andrew Beal猜想:如果A=B=C=1;m;n;r均大于2;则该方程没有正整数解.
It may take a long time to find a solution to the problem. 也许要花很长时间才能找到解决这个问题的办法。
The Positive Integer Solution to a Class of Diophantine Equations 一类丢番图方程的正整数解
Integer Solution to A class of Linear Complementarity Problem 一类线性互补性问题的整数解
I simply refuse to conceive of such a solution. 这样的解决办法，我是绝对不考虑的。
What color is copper sulphate solution? 硫酸铜溶液是什么颜色？
Mary is sceptical about the solution. 玛丽对这个解决办法表示怀疑。
His problem did not really admit of any solution. 他的问题真是没有解决的可能。
He doped out a solution to the problem. 他想出一个解决该问题的方法。