Anik likes division and is always eager to find remainders.
Once Anik divided by to get positive remainder ; and by to get positive remainder .
If Anik assumed as a positive integer greater than 2 and as an positive odd number greater than 1, find .
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We crack this problem using simple congruence:
a ≡ 1 ( m o d a − 1 ) ⇒ a x ≡ 1 ( m o d a − 1 ) ⇒ r 1 = 1
Now we will use the fact that x is odd now:
a ≡ − 1 ( m o d a + 1 ) ⇒ a x ≡ − 1 ≡ a ( m o d a + 1 ) ⇒ r 2 = a
Thus , r 1 + r 2 = a + 1 .