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

Sparticusretiarius

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"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.
 

Subhotosh Khan

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"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?
 
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lookagain

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"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 \(\displaystyle \ N^2\).

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

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


Then we have:

\(\displaystyle 2,000 \ < x^6 \ < 10,000\)

\(\displaystyle \sqrt[6]{2,000} \ < x \ < \sqrt[6]{10,000}\)

\(\displaystyle 3.5... \ < x \ < 4.6...\)


What can you conclude?
 
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JeffM

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It is obvious that

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

It is not obvious, however, that

\(\displaystyle b^2 = c^3 \in \mathbb Z \implies \sqrt{b} \in \mathbb Z.\)

It may be true, but it requires a proof.
 
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