## 2bornot2b 3 years ago 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)

1. 2bornot2b

2. 2bornot2b

@Ishaan94 can you help?

3. FoolForMath

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

4. Mani_Jha

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?

5. Mani_Jha

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

6. 2bornot2b

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

7. 2bornot2b

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

8. Mani_Jha

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?