1：Prove that if x is a positive integer that is not divisible by 5,then x4 -1 is divisibleby 5.（证明如果X是可被5整除的整数,那么X的四次方-1也能被5整除）2：Prove that there do not exist two positive integers x and y such th

1：Prove that if x is a positive integer that is not divisible by 5,then x4 -1 is divisibleby 5.（证明如果X是可被5整除的整数,那么X的四次方-1也能被5整除）2：Prove that there do not exist two positive integers x and y such th
1：
Prove that if x is a positive integer that is not divisible by 5,then x4 -1 is divisible
by 5.
（证明如果X是可被5整除的整数,那么X的四次方-1也能被5整除）
2：
Prove that there do not exist two positive integers x and y such that x2-4y2 = 14.
Additional information:you have all known for a long time that there are many
solutions to the equation x2 + y2 = z2,where x,y and z are all positive integers.
This problem considers a slightly di\x0berent,but similar looking,type of equations.
Hint:use an indirect proof,and start by factoring x2 - 4y2.
（证明不存在两个数使得x方-4y方=14
提示：用间接证明从因式分解x方-4y方开始））
证明结果有可能是真命题也有可能是假命题.
PS：这不是高等数学证明题,只是非数学课学证明方法的课后题,不会用到大二大三那种很高深的数学..
上面翻译错了.
（证明如果X是不可被5整除的整数,那么X的四次方-1可以被5整除）

1：Prove that if x is a positive integer that is not divisible by 5,then x4 -1 is divisibleby 5.（证明如果X是可被5整除的整数,那么X的四次方-1也能被5整除）2：Prove that there do not exist two positive integers x and y such th
1)x^4-1=(x^2-1)(x^2+1)=(x-1)(x+1)(x^2+1)
因为x不能被5整除(你翻错了...)
故有4种可能:
(1)x=5n+1,则由于x^4-1的因子中有x-1,即5n这个因子,故可被5整除
(2)x=5n+2,则由于x^4-1的因子中有x^2+1,即(5n+2)^2+1=25n^2+20n+5,故可被5整除
(3)x=5n+3,则由于x^4-1的因子中有x^2+1,即(5n+3)^2+1=25n^2+30n+10,故可被5整除
(4)x=5n+4,则由于x^4-1的因子中有x+1,即5n+5,故可被5整除
综上获证
2)14=2*7=1*14
x^2-4y^2=(x-2y)(x+2y)
故x要等于2和7的中间数4.5,或1和14的中间数
但两种情况取的x都不是整数(这里你也翻译错了,题目说的是正整数)
故不能取这样的x
故不存在