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How many positive integers n less than 100 have a corresponding integer m divisible by 3 such that the roots of x2nx+m=0  are consecutive positive integers?

 Feb 10, 2019
 #1
avatar+33654 
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Number of positive integers, n, as follows:

 

 Feb 10, 2019
 #2
avatar+63 
+1

Thank you so much!

vindou  Feb 10, 2019

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