"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.
"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 (45^{2} ↔ 100^{2}) - list those
How many integer cubic numbers are there between 2000 ↔ 10000 (13^{3} ↔ 21^{3}) - list those
Hint: How many "square numbers" do you have within 13 ↔ 21?
Last edited by Subhotosh Khan; 02-25-2018 at 08:14 PM.
“... mathematics is only the art of saying the same thing in different words” - B. Russell
"Brute force" was stated, but it is not required.
Because it is a perfect square, it is of the form [tex] \ N^2[/tex].
Because it is also a perfect cube, it is also of the form [tex] \ M^3 [/tex].
[tex] x^6 \ = \ (x^3)^2 \ = \ (x^2)^3 \ [/tex] satisfies both. (The exponent on x is the least common multiple of the other two exponents.)
Then we have:
[tex] 2,000 \ < x^6 \ < 10,000[/tex]
[tex]\sqrt[6]{2,000} \ < x \ < \sqrt[6]{10,000}[/tex]
[tex]3.5... \ < x \ < 4.6...[/tex]
What can you conclude?
Last edited by lookagain; 02-26-2018 at 11:06 AM.
It is obvious that
[tex]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.[/tex]
It is not obvious, however, that
[tex]b^2 = c^3 \in \mathbb Z \implies \sqrt{b} \in \mathbb Z.[/tex]
It may be true, but it requires a proof.
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