Prove the following statement.The square of any odd integer has the form 8m+1 for some integer m.

Step-by-step explanation:As per the question,Let a be any positive integer and b = 4.According to Euclid division lemma , a = 4q + rwhere 0 ≤ r < b.Thus, r = 0, 1, 2, 3Since, a is an odd integer, andThe only valid value of r = 1 and 3So a = 4q + 1 or 4q + 3Case 1 :- When a = 4q + 1On squaring both sides, we geta² = (4q + 1)²    = 16q² + 8q + 1    = 8(2q² + q) + 1    = 8m + 1 , where m = 2q² + qCase 2 :- when a = 4q + 3On squaring both sides, we geta² = (4q + 3)²    = 16q² + 24q + 9    = 8 (2q² + 3q + 1) + 1    = 8m +1, where m = 2q² + 3q +1Now,We can see that at every odd values of r, square of a is in the form of 8m +1.Also we know, a = 4q +1 and 4q +3 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