Explain how to tell if a number is a square number or a cube number or neither?

Question

Please help, it's the last question of my homework and I really want to finish
Thanks

Answer 1

To be honest, there's no super-easy way. The easiest way is to guess-and-check.



For example, say that you have 857, and you want to see if it's a cube number.

9^3 is 729. 10^3 is 1000. Since 857 is between 729 and 1000, it can't be a cube.



Another way to do it might be to try and factor the number into primes, and make sure that each number has a "pair" (or is part of a triplet). For example, take 1936. You know that it's even, so divide by 2 until you can't do it anymore. You get 2 * 2 * 2 * 2 * 121. Now, factor 121, and you get 2 * 2 * 2 * 2 * 11 * 11.



You can pair up 2's and pair up 11's to get:

(2 * 2) * (2 * 2) * (11 * 11)

You can now take one number from each pair and put it in one of two groups, like so:

(2 * 2 * 11) * (2 * 2 * 11)

That's easy to see that it's a square, since both numbers in the parentheses are the same.



Another fun quick trick: All squares end in 0, 1, 4, 5, 6 or 9. So you know that 483957347539873 isn't square.

Answer 2

We know this is true because we can prove it. What does it mean to be a square number? It means that our square root is an integer (and therefore a rational number - a number that can be expressed as a fraction of two integers a,b in the form a/b). However we can prove that the square root of any prime number is irrational (that is, it cannot be expressed as a fraction of integers). Let's prove it: Let p be a prime, assume for contradiction that the square root of p is rational. If sqrt(p) is rational then there exist integers a,b such that sqrt(p)=a/b. Also assume (without loss of generality) that gcd(a,b)=1, that is that a and b are coprime and hence a/b is a fraction in lowest terms. If this is the case then pI(a^2) (p divides (a^2)) which implies pIa. If pIa then there exists some q in the integers such that a=pq, then p(b^2)=(a^2)=(pq)^2=(p^2)(q^2) and therefore b^2=p(q^2). If b^2=p(q^2) then pI(b^2) which implies pIb therefore p is a common factor or both a and b. However a and b are coprime and hence we have a contradiction. Therefore the square root of any prime p is irrational. If the square root of any prime is irrational then no prime can also be a square number.

Answer 3

square number has a 2 above it and a cubed has a three and if its neither then its neither...

Answer 4

put # in calc.. press sqrt... if not integer, put # in calc again and take cubed root... if not integer its neither!

...but im sure thats not what they're asking for :/