Question

show that the square of any positive integer cannot be of the form 5m+2 or 5m+3 for any integer m.

Answer 1

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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