show that the square of any positive integer cannot be of the form 5m+2 or 5m+3 for any integer m.
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Answer:
Let b = 5
r = 0, 1, 2, 3, 4 (0 =< r < b)
When r = 0
a = bq + r
a = 5q + 0
a = 5q
Squaring:
a = 25q^2
a = 5(5q^2) ..(1)
When r = 1
a = 5q + 1
Squaring:
a = (5q + 1)^2
a = 25q^2 + 1 + 10q
a = 5(5q^2 + 2q) + 1 ...(2)
When r = 2
a = 5q + 2
Squaring
a = (5q + 2)^2
a = 25q^2 + 4 + 20q
a = 5(5q^2 + 4q) + 4 ..(3)
When r = 3
a = 5q + 3
Squaring
a = (5q + 3)^2
a = 25q^2 + 9 + 30q
a = 25q^2 + 30q + 4 + 5
a = 5(5q^2 + 6q + 1) + 4 ...(4)
When r = 4
a = 5q + 4
Squaring
a = (5q + 4)^2
a = 25q^2 + 16 + 40q
a = 25q^2 + 15 + 1 + 40q
a = 5(5q^2 + 3 + 8q) + 1 ..(5)
By (1),(2),(3),(4) and (5), we can say that no positive integer can be of the form 5m + 2 or 5m + 3
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