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2015-04-07T20:06:03+05:30

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Let x be an integer.
     x² -1 = (x - 1) (x + 1) 

let  x be an odd positive integer.  x - 1  is an even positive integer or 0 = 2 k (say)
  so x + 1 is then an even positive number = 2 k + 2
          x² - 1 =  4 k (k + 1)
 
Now k and k + 1  are consecutive integers.  Hence one of them is divisible by 2.  If k is even then k+1 is an odd number.  If k is an odd number, then  k+1  is an even number.
   Hence,  the product k (k+1) = 2 * [k/2] * (k+1)  or    k * [ (k+1)/2 ] * 2

   thus  x² -1 =  8 * (k/2) * (k+1)    or  8 * k * [(k+1)/2]

Hence,    x² - 1  is divisible by 8.
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you can use the method of proof by induction.

x² - 1  is divisible by  8  if x = 1 .

let x² - 1  be divisible by 8  for some x.  Now let us see for the next off number  x+2.
   (x+2)² - 1
        =  x² + 4x + 4 - 1
        =  (x² - 1) + 4 (x + 1)
        as x is an odd number,  x +1 is an even number.  Hence 4 (x+1) is divisible by 8.
   Since both terms on RHS are divisible by 8.,  then the (x+2)² - 1 is divisible by 8.

Hence, proved by mathematical induction method.

2 5 2
the essence or summary of the proof is same , when you use the lemma of Euclid. Only thing is to use the language and symbols used in the lemma.
p | a means that p divides a.
Euclid's lemma says that p | a*b, then p | a or p | b or p | a*b.
ALternately, if p |a, then p| ab. --- (5)
Also, we can say that p | a/b, then pb | a. --- (6)
Let x = 2 m + 1, as any odd integer is expressed in this manner. m is an integer >= 0.
x² - 1 = (2 m + 1)² - 1 = 4 m (m + 1) --- (1)
m(m-1) = (x² - 1)/4 = an integer ---- (2)
We know that 2 | m (m+1), as one of m and m+1 is an even integer. They are consecutive integers. that is,
either 2 | m or 2 | (m+1) , hence, as per the Lemma - (5), 2 | m * (m+1)
m(m+1) | (x²-1)/4 --- (3)
Since, 2 | m(m+1) => 2 | (x² - 1)/4
=> 8 | (x²-1) by using -- (6)
hence, the proof is done.