平方和四次方的比较有什么窍门么 谢谢

bernadotte

我不会啊 请见图

bernadotte

still 4, which are 3,5,7,11.

9,25,49,121 is not prime number.

-- by 会员 liyang229 (2010/1/27 4:11:34)

说的也是

many thanks!

顺便再帮我看看那个x^4+y^4>z^4 和它们的平方的和的比较吧

liyang229

still 4, which are 3,5,7,11.

9,25,49,121 is not prime number.

bernadotte

for statement 2

it also could be k =3^2*5^2*7^2*11^2
then it becomes a square of some number.

-- by 会员 liyang229 (2010/1/27 4:00:07)

如果那样的话 岂不是有8个质数了 题目说正好4个啊?

liyang229

for statement 2

it also could be k =3^2*5^2*7^2*11^2
then it becomes a square of some number.

bernadotte

for statement 2, we could know see k =3*5*7*11 or whatever multiple for 4 prime number.
so, it cannot be a square of number.
====
那不就正好解释了K不是整数的平方了么

liyang229

answer is E.
for statement 1, if k is equal to 8. it is not a square of integer.
if K= 4, it is . so, statement 1 is not sufficient enough.

for statement 2, we could know see k =3*5*7*11 or whatever multiple for 4 prime number.
so, it cannot be a square of number.

so statement 2 is also wrong.

answer is E