<PRE><B>Question 63910</B>
Find two consecutive positive numbers 
N AND N+1 SAY
such that the product of the sum(N+N+1=2N+1) and difference(N+1-N=1) of the numbers plus eight 
(2N+1)1+8
is the sum of their squares[(N^2+(N+1)^2].
2N+1+8=N^2+(N+1)^2=N^2+N^2+1+2N
2N^2-8=0
N^2=8/2=4
N=+2 0R -2
HENCE THE 2 NUMBERS ARE 
2 AND 3 .....CHECK...(2+3)(3-2)+8=2^2+3^2.....13=13..OK
OR
-2 AND -1..........CHECK.....(-2-1)(-1+2)+8=(-2)^2+(-1)^2...OR....5=5..OK





THANK YOU!!!!!!
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