SOLUTION: Recall that an integer d is said to be a divisor of an integer a if a/d is also an integer. For how many integers a between -200 and -1 inclusive is the product of the divisors of

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Question 1210675: Recall that an integer d is said to be a divisor of an integer a if a/d is also an integer. For how many integers a between -200 and -1 inclusive is the product of the divisors of a negative? It is not 0.
Answer by ikleyn(53979) About Me  (Show Source):
You can put this solution on YOUR website!
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Recall that an integer d is said to be a divisor of an integer a if a/d is also an integer.
For how many integers a between -200 and -1 inclusive is the product of the divisors of a negative?
It is not 0.
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Let N be an integer between -200 and -1, -200 <= N <= -1.


Consider the absolute value |N|, the opposite integer to N, which is a positive integer number.

Let

    |N| = %28p%5B1%5D%29%5En%5B1%5D%2A%28p%5B2%5D%29%5En%5B2%5D%2Aellipsis%2A%28p%5Bm%5D%29%5En%5Bm%5D

be the decomposition of number |N| into the product of prime factors p%5B1%5D, p%5B2%5D, . . . , p%5Bm%5D.


Then the number of all positive divisors of |N| is %281%2Bn%5B1%5D%29%2A%281%2Bn%5B2%5D%29%2Aellipsis%2A%281%2Bn%5Bm%5D%29.


It is a well-known fact from number theory.


The number of all possible negative divisors of the negative integer N is %281%2Bn%5B1%5D%29%2A%281%2Bn%5B2%5D%29%2Aellipsis%2A%281%2Bn%5Bm%5D%29.


The product of all negative divisors of N is a negative number if and only if the number of its divisors is 
an odd number.


Thus, the product of all negative divisors of N is a negative number if and only if
the product %281%2Bn%5B1%5D%29%2A%281%2Bn%5B2%5D%29%2Aellipsis%2A%281%2Bn%5Bm%5D%29 is an odd number.


In turn, the product %281%2Bn%5B1%5D%29%2A%281%2Bn%5B2%5D%29%2Aellipsis%2A%281%2Bn%5Bm%5D%29 is an odd number if and only if
each factor %281%2Bn%5B1%5D%29, %281%2Bn%5B2%5D%29, . . . , %281%2Bn%5Bm%5D%29 is an odd number.


It is so if and only if each index / degree n%5B1%5D, n%5B2%5D, . . . , n%5Bm%5D is an even number.


It happens if and only if the number N is a negative square.


The negative squares between -1 and -200 are -1, -4, -9, -16, -25, -36, -49, -64, -81, -100, -121, -144, -169, -196.


In all, there are 14 such negative squares.


So, the answer to the problem question is 14.

Solved.


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    To the person who posted this problem,
        to the person who created this post/problem (to the composer),
            and to readers of this post.

I have a fundamental professional Math education from a renowned university.
I have read tons of mathematical literature, from popular Math books through
textbooks of all possible levels to professional articles and monographs.
As such, I say that the formulation of this problem in the incoming post is not a professional:
there remain many questions for the reader.
Therefore, below is my reformulation of how the problem should be. 

    Let's consider negative integer numbers { N } between -200 and -1, inclusive.
    For every such number N, let D(N) be the set of all its negative integer divisors
    (each such set includes -1 and N itself).
    For every such number N, let P(N) be the product of all integers of the set D(N).
    We call a number N favorable if the number P(N) is negative.
    How many favorable negative integers N are there in the interval [-200,-1]?

For you to better understand the meaning of the problem and the idea of the solution,
I'll say that it is equivalent to this question: 

    Describe all natural numbers from 1 to 200 inclusive
    that have an odd number of divisors.