Question

Prove that hte square of anu postive intgers cannot be of the form 5q+2 or 5q+3 where q is some integer.

Answer 1

Answer:

5q+2 must end in 2 (if q is off) or 7 (if q is even) as the Least Significant Digit.

5q+3 must end in 3 or 8.

Any integer p must have its LSD = the LSD of the square of its LSD. (e.g. the LSD of the square of 8327 is the same as the LSD of the square of 7 i.e. 9 in this example.)

By exhaustive enumeration of 0^2, 1^2 .. 8^2, 9^2, it can be verified the LSD in each case is in {0, 1, 4, 5, 6, 9}. So no square of an integer can end in 2, 3, 7 or 8.

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Answer 2

Step-by-step explanation:

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Let a be an arbitrary positive integer.Then, by Euclid ’s divisions Algorithm, corresponding to the positive integers a and 5, there exist non-negative integers m and r such thata = 5m + r, where 0 ≤ r < 5⇒ a2 = (5m + r2) = 25mr + r2 + 10mr [∵(a+b)2 = a2 + 2ab + b2]⇒ a2 = 5(5m2 + 2mr) + r2 ...(i)where, 0 ≤ r < 5