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Perfect Square, Cube, Fourth Power


Date: 01/25/2002 at 19:05:54
From: James A. Hopton
Subject: Perfect square, cube, fourth power

My wife is an 8th Grade Math Teacher, and in the Bonus Section in 
one of the chapters of the current math book is this problem:

Find the least integer greater than 1 that is a perfect square, a 
perfect cube, and a perfect fourth power.

There was no equation with the problem. We have come up with the 
number 4096, (do not know if this is correct), by pushing and shoving 
numbers around. Is there an equation that will enable someone to solve 
this problem in a reasonable amount of time?

Thank you for your help, and we anxiously await your answer.


Date: 01/25/2002 at 20:06:08
From: Doctor Paul
Subject: Re: Perfect square, cube, fourth power

Every number can be written as a product of powers of primes - this is 
the Fundamental Theorem of Arithmetic.

A number is a perfect square if, when raised to the 1/2 power, it 
still has an integer for all of the exponents in its prime 
factorization.

For example:

3136 = 2^6*7^2 is a perfect square because 
3136^(1/2) = [2^6*7^2]^(1/2) = 2^3*7 = 56.

Similarly, a number will be a cube root if, when raised to the 1/3 
power, it still has an integer for all of the exponents in its prime 
factorization.

A number will be both a perfect square and a perfect cube if, when 
raised to both the 1/2 and the 1/3 powers, it still has an integer for 
all of the exponents in its prime factorization.  otice that 
successively raising a number to the 1/2 and then the 1/3 power is the 
same as taking the 6th root of a number. So the numbers that are going 
to be perfect squares and perfect cubes are precisely the 6th powers 
of the integers:

1^6 = 1
2^6 = 64
3^6 = 729
4^6 = 4096
5^6 = 15625
etc...

A similar computation shows that for a number to be a perfect square, 
a perfect cube, and the fourth power of some integer, it must be the 
12th power of some integer. Note that we don't require the 24th power.  
The 24th power will work, but it's overkill. Essentially, you want to 
compute lcm(2,3,4) = lcm(2, lcm(3,4) ) = lcm(2,12) = 12. Does this 
make sense?

Notice that the square root of x^12 = x^6, the cube root of x^12 is 
x^4, and the 4th root of x^12 is x^3.

So the numbers that will be perfect squares, cubes, and fourth powers 
are the powers of 12:

1^12 = 1
2^12 = 4096
3^12 = 531441

etc...

You have correctly computed (I guess by trial and error) that 4096 is 
the right answer.  Perhaps now you understand why.

I hope this helps. Please write back if you'd like to talk about this 
more.

- Doctor Paul, The Math Forum
  http://mathforum.org/dr.math/   
    
Associated Topics:
High School Number Theory

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