the question is based on number system?

Question

prove that the number which is simultaneously a perfect square and perfect cube is always in the form of 7n+1 or 7n-1

Answer 1

Work modulo 7, i.e., with remainders when you divide by 7. What are the possible squares? 0, 1, 4, 2 (since 3^2 = 9 which is congruent to 2 mod 7), and the other squares are already on the list. What are the possible cubes? 0, 1, 6. So a number that both a square and a cube looks like it has to be of the form 7n or 7n+1.

Answer 2

So x = a^6 for some integer a.

(Yes, a has to be an integer because all the exponents in the prime factoring of x must be multiples of 2 and of 3 for x to be a square and a cube. Therefore they are all multiples of 6 and x is indeed a sixth power.)

Now a = 7n + r, where r = 0, 1, 2, 3, 4, 5 or 6.

Then x = a^6 = (7n+r)^6
= (7n)^6 + 6(7n)^5(r) + (binomial expansion)... + 6(7n)r^5 + r^6
= r^6 (modulo 7) (all other terms are multiples of 7).

Now 0^6 = 0 (mod 7)
1^6 = 1 (mod 7)
2^6 = 64 = 1 (mod 7)
3^6 = 729 = 1 (mod 7)
4^6 = 4096 = 1 (mod 7)
5^6 = 15625 = 1 (mod 7)
6^6 = 46656 = 1 (mod 7)

So the remainder of x divided by 7 will always be 0 (if a is a multiple of 7) or 1. x is of the form 7n or 7n+1.

7n-1 is neither here nor there.

Answer 3

First of all, your statement is not quite accurate.
Let m be your number. If m is a multiple of 7,
it's square and cube are both of the form 7n.
Next, I'll prove a stronger statement:
The cube of any number not a multiple of 7
is always of the form 7n+1 or 7n-1.
(So m doesn't have to be a square.)
To see this, recall that any m not a multiple 7 of can be written in the form
7q + k, with k = 1,2,3,4,5, or 6.
Also note that (7q+k)³ = 343q³ +147q²k +21qk² + k³.
= 7n + k³.
So all we have to do is look at k³ for k = 1,2,3,4,5,6.
k=1: 1³ = 7*0+1
k=2: 2³ = 7*1+1
k=3: 3³ = 7*3+6 = 7*4-1
k=4: 4³ = 7*9+1
k=5: 5³ = 7*17+6 = 7*18-1
k=6: 6³ = 7*30+6 = 7*31-1
So we see that any cube not a multiple of 7
is always of the form 7n+1 or 7n-1.
BTW: The same result is true if you replace
7 by 9 and let m be any number not a
multiple of 3.