2 − 1 1 − 3 − 2 1 + 4 − 3 1 − 5 − 4 1 + ⋯ + 6 4 0 0 − 6 3 9 9 1 = n 2
Determine ∣ n ∣ .
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2 − 1 1 − 3 − 2 1 + 4 − 3 1 . . . + 6 4 0 0 − 6 3 9 9 1 =
2 − 1 2 + 1 − 3 − 2 3 + 2 + 4 − 3 4 + 3 . . . + 6 4 0 0 − 6 3 9 9 6 4 0 0 + 6 3 9 9
Now that we have a common denominator of 1 , we can ignore the denominator. Notice that all the terms cancel each other out except for the two end terms, 6 4 0 0 and 1 . Thus,
n 2 = 8 1 ⇒ ∣ n ∣ = 9
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rationalise the denominator. On rationalising you will find that all terms cancel off except sq.rt(1) and sq.rt(6400). So on adding we find that n^2 =81 mod(n)=9