Have You Tried Squaring A Square?

Number Theory Level 1

How many integer(s) are there that is not a perfect square , but a perfect cube, and a perfect fourth power?

0 4 8 2

3 solutions

Zee Ell
Aug 27, 2016

All 4th powers are perfect squares (square of a square):

$n^4 = (n^2)^2$

Therefore, there are no such integers which are perfect 4th powers, but not perfect squares at the same time.

I liked Kushal Bose's solution,but yours is simple and clear.Me thought that too:)

Anandmay Patel - 4 years, 9 months ago

Kushal Bose
Aug 26, 2016

Let $N$ be such a number.

So, $N=a^3=b^4$ where $a,b \in \mathbb I$

$a^3=b^4$

$a=b^{4/3}$

$a=b^{1+1/3}$

$a=b \sqrt[3]{b}$

Let, $b=k^3$ where $k$ is an integer

$a=k^3.k=k^4$

Putting this in the first equation :

$N=a^3=b^4=k^{12}$

So , $N$ becomes a perfect square.

No solution exists

How can you let that b=k^3? (where k is necessarily an integer?)

Anandmay Patel - 4 years, 9 months ago

because cube root of b should be an integer

Kushal Bose - 4 years, 9 months ago

But see your second line.According to it,the 4th root of b should be an integer.[??]

Anandmay Patel - 4 years, 9 months ago

@Anandmay Patel – Can you tell me in detail which portion you are facing problem ?

Kushal Bose - 4 years, 9 months ago

@Kushal Bose – Thanks for attention.You specified that N=b^4 for some integer b.BUT how is b a perfect cube? You assumed the value of b to be equal to the cube of another integer k.This means that b is a perfect cube.BUT how can you say that[that b is a perfect cube?]?Any proof?

Anandmay Patel - 4 years, 9 months ago

@Anandmay Patel – Do you agree with this line :

$a=b \sqrt[3]{b}$

If u agree then follow this:

In L.H.S. a is an integer then R.H.S. should also be an integer .Then $\sqrt[3]{b}$ should also be an integer.This is why I have assumed that b is a perfect cube.

Kushal Bose - 4 years, 9 months ago

@Kushal Bose – Gotit!Thanks!

Anandmay Patel - 4 years, 9 months ago

@Kushal Bose – Will you please give me an example of a number having the property specified in the question,but this time,it also is a perfect square?Thanks.

Anandmay Patel - 4 years, 9 months ago

Noel Lo
Aug 29, 2016

Seriously...a perfect fourth power is also a perfect square, isn't it?

yes $\quad$ $\quad$ $\quad$ $\quad$ $\quad$

Pi Han Goh - 4 years, 9 months ago

Have You Tried Squaring A Square?

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