show the sq of any positive integer is of the form 4m or 4m +1 where m is any integer
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We know that any positive integer can be of the form 4m or 4m + 1, for some integer m.
Thus, anypositive integer can be of the form 4q or 4q + 1.
Thus we have (4q)2 = 16q2 = 4(4q2) = 4m where m = 4q2 (4q+1)2 = 16q2 + 8q +1 = 4(4q2 +2q) +1 = 4m + 1 where m = 4q2 +2q.∴ Square of any positive integer is of the form 4m or 4m+1, where m is any integer
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We know that any positive integer can be of the form 4m or 4m + 1, for some integer m.
Thus, anypositive integer can be of the form 4q or 4q + 1.
Thus we have (4q)2 = 16q2 = 4(4q2) = 4m where m = 4q2 (4q+1)2 = 16q2 + 8q +1 = 4(4q2 +2q) +1 = 4m + 1 where m = 4q2 +2q.∴ Square of any positive integer is of the form 4m or 4m+1, where m is any integer
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We know that any positive integer can be of the form 4m or 4m + 1, for some integer m.
Thus, anypositive integer can be of the form 4q or 4q + 1.
Thus we have (4q)2 = 16q2 = 4(4q2) = 4m where m = 4q2 (4q+1)2 = 16q2 + 8q +1 = 4(4q2 +2q) +1 = 4m + 1 where m = 4q2 +2q. ∴ Square of any positive integer is of the form 4m or 4m+1, where m is any integer
Thus, anypositive integer can be of the form 4q or 4q + 1.
Thus we have (4q)2 = 16q2 = 4(4q2) = 4m where m = 4q2 (4q+1)2 = 16q2 + 8q +1 = 4(4q2 +2q) +1 = 4m + 1 where m = 4q2 +2q. ∴ Square of any positive integer is of the form 4m or 4m+1, where m is any integer
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srry for the wrong answer but here is the correct answer...
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