SOLUTION: THE SUM OF THE SQUARES OF TWO CONSECUTIVE POSITIVE INTEGERS IS 41 41=n^2(n+1)^2 n^2+n^2+2n-40=0 n^2+2n-40=0 factor out 2 n^2+n-20=0 I tried different factors of -20 (n+1)(n-2

Question 394704: THE SUM OF THE SQUARES OF TWO CONSECUTIVE POSITIVE INTEGERS IS 41
41=n^2(n+1)^2
n^2+n^2+2n-40=0
n^2+2n-40=0
factor out 2
n^2+n-20=0
I tried different factors of -20 (n+1)(n-20)=-19; (n+2)(n-10)=-8
(n+10)(n-2)=8 (n+20)(n-1)=19 i know the answers are supposed to be 2 and 4 so I know I am wrong but where?
Answer by richard1234(7193) (Show Source): You can put this solution on YOUR website!
-20 is also equal to 4*(-5) or (-4)*5. You should factor it as

in which the values of n are 4 and -5. Since the integers are positive, n = 4.
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