Question

If a number is both square and cube, show that it is of the form $7n$ or $7n+1$.

Answer 1

Every positive integer which is both square cube

of a +ve integer must be a 6th power of a

positive integer

we will show that for energy n,

$n^6$ is either in the form of $7k$ or $7k+1$

integer can be writer as

$7p+q$

if $q=0$ then

$(7-p{)}^6=7^6{p}^6$ which is multiple of 7

For other case

$n^6=7k+1$

which will be true if $7k=n^6-1$

So if we show that for every n, are of these four

following factor is multiple of 7

$n^6-1=(n-1)(n^2+n+1)(n+1)(n^2-n+1)$

Suppose $n=7p+q$

then i)$x-1=7p+q-1$ multiple of seven if $q=1$

$ii)n^2+n+1=(7p+q{)}^2+(7p+q)-1=49p^2+1+1q+7p+q^2+q+1$

is a multiple of T off $(q^2+q+1)$ is a multiple 87

iii) $n+1=7p+q+1$ is multiple of 7 if q = 6

iv) $n^2-n+1=49p^2+1+pq-7p+q^2-q+1$ if multiple of 7

$q^2-q+1$ is multiple of 7

when q= 1 (i) is multiple of 7 q =5 (iv)is multiple of 7

q = 2 (ii) is multiple of 7 q = 6 (iii) is multiple of 7

q = 3 (iv) is multiple of 7

q = 4 (ii) is multiple of 7

Hence it is proved