Question

Prove that if an integer is a perfect square and a perfect cube, simultaneously, then it is either of the form 7k or 7k+1
(The problem belongs to the chapter, division algorithm)

Answer 1

@Mani Jha @Zarkon @FoolForMath @JamesJ Please help!

Answer 2

@Ishaan94 can you help?

Answer 3

Standard number theory problem. This should help : http://www.math.wisc.edu/~jensen/567/hwk1.pdf

Answer 4

If an integer is both a square and a cube, it can be of the form:
\[(a ^{3})^{2}\]
Now, since a cube can be of the form 7k or 7k+-1(thanks to FoolForMath), we write
\[a ^{3}=7k\]
and get the no to be 49k^2, which is in the form of 7 times something
\[49k ^{2}=7\times(7k ^{2})\]
Now put
\[a ^{3}=7k+-1\]
Square it and you'll get a number in the form of (7times something +1)
Do you want me to show the steps that I skipped?

Answer 5

Luis Rivera and I are the only people here who have usernames with a space. So, we can't be pinged :(

Answer 6

@FoolForMath can you provide me a similarn pdf containing problem and solution set for topics related to divisibility, prime numbers, and Greatest Common divisor?

Answer 7

I didn't get this line "If an integer is both a square and a cube, it can be of the form:
(a^3)^2" written by @Mani Jha

Answer 8

Consider 64. It is the cube of 4 and square of 8. It can be written as:
8^2=(2^3)^2
I've assumed that this is valid for all numbers which are powers of 6(64=2^6) or of multiples of 6.
\[a ^{6}=(a ^{3})^{2}\]
Any other power like a^2 a^3 a^4 can't be both a square and cube of some integer.
Clear?