设自然数n1>n2,且n12-n22=79,则n1=_,n2=_.

问题描述:

设自然数n1>n2,且n12-n22=79,则n1=______,n2=______.

∵n12-n22=79,即(n1-n2)(n1+n2)=79,79=1×79,
∴n1-n2=1,n1+n2=79,
∴n1=40,n2=39.
故答案为40,39.