设自然数n1>n2,且n1^2-n2^2=79,则n1=_____,n2=____

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设自然数n1>n2,且n1^2-n2^2=79,则n1=_____,n2=____
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设自然数n1>n2,且n1^2-n2^2=79,则n1=_____,n2=____
设自然数n1>n2,且n1^2-n2^2=79,则n1=_____,n2=____

设自然数n1>n2,且n1^2-n2^2=79,则n1=_____,n2=____
n1^2-n2^2=79
(n1+n2)(n1-n2)=79=79*1
n1+n2=79
n1-n2=1
n1=40
n2=39

(n1+n2)(n1-n2)=79
因为自然数n1>n2,79为质数,
所以n1+n2=79
n1-n2=1
解得n1=40 n2=39