一道英文数学题..if a<b,then 3^2+4^2+5^2+12^2=a^2+b^2 (符号^是平方) is satisfied by only one pair of positive intergers (a,b).what is the value of a+b?

问题描述:

一道英文数学题..
if a<b,then 3^2+4^2+5^2+12^2=a^2+b^2 (符号^是平方) is satisfied by only one pair of positive intergers (a,b).what is the value of a+b?

问题的意思是方程3方+4方+5方+12方=a方+b方当a

3^2+4^2+5^2+12^2=a^2+b^2
5^2+13^2=a^2+b^2
a+b=5+13=18(because(a,b)is a positive pair)

3^2+4^2+5^2+12^2=194
a^2+b^2=194
那么a=5,b=13
3^2+4^2=25,可以是5的平方
5^2+12^2可以是13的平方
那么the value of a+b,a+b的结果就是13+5=18