Question 1030558: verify that if an integer is simultaneously a square and a cube then it must beveither of the form 7k or 7k+1.
Found 2 solutions by richard1234, Edwin McCravy:
Answer by richard1234(7193)   (Show Source):
Answer by Edwin McCravy(20059)  (Show Source):
You can put this solution on YOUR website!
Here is a way to prove it without calling on advanced
algebraic or number theory theorems, just basic algebra.
Every positive integer which is both a square and a cube
of a positive integer is a 6th power of a positive integer.
We will show that for every n,
Theorem: n6 is either of the form 7k or 7k+1
Every integer can be written in the form:
7p+q where q=0,...6
If q=0, then the theorem is immediate since
(7p)6 = 76p6 which is a multiple of 7.
For the other cases we want to show that n6 = 7k+1,
which will be true if and only if n6-1 = 7k
n6-1 = (n3-1)(n3+1) = (n-1)(n2+n+1)(n+1)(n2-n+1)
If we can show that for every n, one of those four factors is a
multiple of 7, the theorem will be proved.
Suppose n = 7p+q, then substituting in each of the four factors
(1) n-1 = (7p+q)-1 = 7p+q-1, a multiple of 7 when q=1
(2) n2+n+1 = (7p+q)2+(7p+q)+1 = 49p2+14pq+7p+q2+q+1,
a multiple of 7 iff q2+q+1 is a multiple of 7
(3) n+1 = (7p+q)+1 = 7p+q+1, a multiple of 7 when q=6
(4) n2-n+1 = (7p+q)2-(7p+q)+1 = 49p2+14pq-7p+q2-q+1,
a multiple of 7 iff q2-q+1 is a multiple of 7
when q=1, (1) is a multiple of 7
when q=2, (2) is a multiple of 7 because q2+q+1 = 22+2+1 = 7
when q=3, (4) is a multiple of 7 because q2-q+1 = 32-3+1 = 7
when q=4, (2) is a multiple of 7 because q2+q+1 = 42+4+1 = 21
when q=5, (4) is a multiple of 7 because q2-q+1 = 52-5+1 = 21
when q=6, (3) is a multiple of 7
The theorem is proved.
Edwin
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