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How many numbers x between 1 and 100 (inclusive) have the property that x2+x3 is a perfect square?

(Please help. I'm suffering)

User Adam Neal
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The both the question and first answer is crazy confusing so I am simply gonna pray u succeed ma boi
User Andykiteman
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Answer: 9

=========================================================

Step-by-step explanation:

Factor the given expression

x^2+x^3

x^2*1+x^2*x

x^2(1+x)

x^2(x+1)

We end up with something where the factor x^2 is a perfect square regardless of whatever we replace x with from the set {1,2,3,...,99,100}.

In short, we don't need to worry about the x^2 portion.

--------------------

The only thing of importance is the x+1 factor.

The x+1 must be a perfect square so that the original expression is as well.

Let's say M is some perfect square between 1 and 100. That means M belongs to the set of values

{1, 4, 9, 16, 25, 36, 49, 64, 81, 100}

If we want M+1 to be a perfect square, then we'll need to subtract 1 from each item in that set above to counteract the "plus 1" in "M+1".

Subtract 1 from each item to end up with this modified set

{0, 3, 8, 15, 24, 35, 48, 63, 80, 99}

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Every item mentioned in that second set will make x^2+x^3 to be a perfect square.

For example, let's pick x = 15

x^2(x+1) = 15^2*(15+1)

x^2(x+1) = 15^2*16

x^2(x+1) = 15^2*4^2

x^2(x+1) = (15*4)^2

x^2(x+1) = 60^2

This confirms that x^2+x^3 is a perfect square for x = 15. I'll let you check the other values.

There are 10 values in the set {1^2,2^2,...,9^2,10^2} = {1,2,...,81,100}. There are also 10 values in the set {0,3,8, ..., 80, 99} we got earlier. The cardinality hasn't changed. However, we'll kick out 0 because it's not between 1 and 100. So we end up with 9 as our final answer.

--------------------

To summarize, there are 9 values of x that make x^2+x^3 a perfect square, where x is some whole number between 1 and 100 inclusive.

User Jbarradas
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