How do you tell whether a number is a perfect square from the exponents of its prime factorization? Explain why this works.

if it is a perfect square, all of its prime powers will be even.

Since if you have

(2^a 3^b 5^c ...)^2

all the exponents are doubled.

I can tell this works because all the exponents are even, all the exponents are doubled like in a perfect square: 2 ² = 4 and 9 ² = 81.

Why does this work though???

i go yo som

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I'm sorry, but it seems like this is not related to the previous question about perfect squares. Is there something else I can assist you with?

To determine whether a number is a perfect square from the exponents of its prime factorization, you need to check that all the exponents in the prime factorization are even. If all the exponents are even, the number is a perfect square. Otherwise, it is not.

1. Prime factorize the number: Start by finding the prime factors of the given number. This can be done by dividing the number by prime numbers starting from 2 until you can no longer divide.

2. Write down the prime factorization: Express the number as a product of its prime factors. For example, if we take the number 36, its prime factorization is 2^2 * 3^2.

3. Check the exponents: Look at the exponents next to the prime factors. If all the exponents are even (i.e., divisible by 2), then the number is a perfect square. In our example, both exponents are 2, which is even, confirming that 36 is a perfect square.

Why does this work?
When you represent a number as a product of its prime factors, the exponents indicate the number of times each prime factor is multiplied. In a perfect square, each prime factor occurs in pairs, since multiplying a number by itself results in an even exponent for that prime factor. Therefore, if every prime factor has an even exponent, the number is a perfect square. If any prime factor exponent is odd, there will be an unmatched factor, indicating that the number is not a perfect square.