Question 1069384
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We will use the theorem:

If a perfect square is divisible by a prime p,
it is also divisible by p².

Assume that when we divide n²-2 by 4
we get an integer k

{{{(n^2-2)/4}}}{{{""=""}}}{{{k}}}

{{{n^2-2}}}{{{""=""}}}{{{4k}}}

{{{n^2}}}{{{""=""}}}{{{4k+2}}}

{{{n^2}}}{{{""=""}}}{{{2(2k+1))}}}

Therefore n² is divisible by 2.

Since n² is divisible by 2, and 2 is a prime,
by the theorem n² must be divisible by 2², or 4. 

Therefore 2k+1 must be divisible by 2, 
but 2k+1 is an odd number and is not divisible
by 2, so we have reached a contradiction.

Therefore the assumption that n²-2 is divisible by 4
is incorrect, and therefore n²-2 is not divisible by 4.

Edwin</pre></b></font>