Problem 9

Part (a):

Ne are asked to show that the square of an even integer leaves a remainder of 0 when divided by 4.

Solution:

Let n be an even integer. Therefore, n can be written as:

n = 2k for some integer k.

Now, squaring n:

ni = （27）3=4%3.

Clearly, n? is divisible by 4, since n? = 4k2. Therefore, when the square of an even

integer is divided by 4, it leaves a remainder of 0.

Part (b):

Ne are asked to show that the square of an odd positive integer leaves a remainder of 1 when divided by 4, using the diagram or otherwise.

Solution:

Let n be an odd integer. Thus, n can be written as:

n = 2k + 1 for some integer k.

Now, squaring n:

n2= （2k +1）2= 4RP+4k+1=4（k2+k）+1.

So, n? is of the form 4m + 1, where m = k? + k. This means when n? is divided by

4, the remainder is 1. Part (c):

This part refers to the concept of "identisquares" - numbers whose digits are all the same. We are asked to show that there are no such squares with more than one digit, and to verify that there are four single-digit identisquares: 0, 1, 4, and 9.

Solution:

An "identisquare" is a square number where all digits of the number are the same. We are given four single-digit squares: 0, 1, 4, and 9. The task is to prove that no square of a number with two or more digits has all identical digits.

Consider the possible identisquares with two or more digits. The only candidates could be numbers like 11, 22, 33, etc. However, if we square these numbers, we observe that none of them result in a number with identical digits:

• 112 = 121

222 = 484

332 = 1089

442 = 1936

552 = 3025

662 = 4356

77* = 5929

882 = 7744

992 = 9801 None of these squares have all identical digits. Therefore, there are no identisquares with more than one digit.