Question

# How to prove that the square of an odd integer can express in the form of (8m+1), for some positive integer m?

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Solution

## According to Euclid division lemma , a = bq + r where 0 ≤ r < b Here we assume b = 8 and r ∈ [1, 7 ] means r = 1, 2, 3, .....7 Then, a = 8q + r Case 1 :- when r = 1 , a = 8q + 1 squaring both sides, a² = (8q + 1)² = 64²q² + 16q + 1 = 8(8q² + 2q) + 1 = 8m + 1 , where m = 8q² + 2q case 2 :- when r = 2 , a = 8q + 2 squaring both sides, a² = (8q + 2)² = 64q² + 32q + 4 ≠ 8m +1 [ means when r is an even number it is not in the form of 8m + 1 ] Case 3 :- when r = 3 , a = 8q + 3 squaring both sides, a² = (8q + 3)² = 64q² + 48q + 9 = 8(8q² + 6q + 1) + 1 = 8m + 1 , where m = 8q² + 6q + 1 You can see that at every odd values of r square of a is in the form of 8m +1 But at every even Values of r square of a isn't in the form of 8m +1 . Also we know, a = 8q +1 , 8q +3 , 8q + 5 , 9q +7 are not divisible by 2 means these all numbers are odd numbers Hence , it is clear that square of an odd positive is in form of 8m +1

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