If p is a positive odd integer, what is the remainder when p is divided by 4 ?
(1) When p is divided by 8, the remainder is 5.
(2) p is the sum of the squares of two positive integers.
Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient.
Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient.
BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient.
EACH statement ALONE is sufficient.
Statements (1) and (2) TOGETHER are NOT sufficient.
(1)余数必为1
(2)令p=a方+b方。 p是奇数,则a,b一奇一偶。假设a为偶数,则a必有因数2,从而a方必有因素4,从而a方能被4整除,p/4的余数=b方/4的余数。用数学语言表达b=2n+1, (n大于等于0)。因此b方=4n方+4n+1,除以4余数是1,因此p/4的余数是1。
(2)令p=a方+b方。 p是奇数,则a,b一奇一偶。假设a为偶数,则a必有因数2,从而a方必有因素4,从而a方能被4整除,p/4的余数=b方/4的余数。用数学语言表达b=2n+1, (n大于等于0)。因此b方=4n方+4n+1,除以4余数是1,因此p/4的余数是1。
做②的时候漏看了P是奇数的条件 2)p必定是1奇和1偶数的平方之和
(1) p/8=a+5-->p=8a+5, p/4=(8a+5)/4=2a+5/4=3a+1,无论p是奇是偶,p/4余数是1.
漏看了条件
作②的时候漏看了P是奇数的条件
条件2:a2+b2=p奇数,那么a奇b偶,a=2n+1,b=2m,则4n^2+4n+1+4m^2=p, p除以4只能余1
因为p是奇数 所以不可能是两个相同的数的平方之和,也不可能是两个偶数平方之和
条件2:a2+b2=p奇数,那么a奇b偶,a=2n+1,b=2m,则4n^2+4n+1+4m^2=p, p除以4只能余1
第二个条件用数字试