Thread: Find a whole number that is both perfect square and cube found between 2000 and 10000

1. Find a whole number that is both perfect square and cube found between 2000 and 10000

"Find a whole number that is both perfect square and cube found between 2000 and 10000?"

I am kind of stumped on this question. If anyone can help me that would great.

2. Originally Posted by Sparticusretiarius
"Find a whole number that is both perfect square and cube found between 2000 and 10000?"

I am kind of stumped on this question. If anyone can help me that would great.
Brute Force!!

How many integer square numbers are there between 2000 ↔ 10000 (452 ↔ 1002) - list those

How many integer cubic numbers are there between 2000 ↔ 10000 (133 ↔ 213) - list those

Hint: How many "square numbers" do you have within 13 ↔ 21?

3. Originally Posted by Sparticusretiarius
"Find a whole number that is both perfect square and cube found between 2000 and 10000?"

"Brute force" was stated, but it is not required.

Because it is a perfect square, it is of the form $\ N^2$.

Because it is also a perfect cube, it is also of the form $\ M^3$.

$x^6 \ = \ (x^3)^2 \ = \ (x^2)^3 \$ satisfies both. (The exponent on x is the least common multiple of the other two exponents.)

Then we have:

$2,000 \ < x^6 \ < 10,000$

$\sqrt[6]{2,000} \ < x \ < \sqrt[6]{10,000}$

$3.5... \ < x \ < 4.6...$

What can you conclude?

4. It is obvious that

$a \in \mathbb Z \implies a^2,\ a^3, a^6 \in \mathbb Z \text { and } (a^2)^3 \equiv a^6 \equiv (a^3)^2.$

It is not obvious, however, that

$b^2 = c^3 \in \mathbb Z \implies \sqrt{b} \in \mathbb Z.$

It may be true, but it requires a proof.

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