Let's review and correct the solution step by step for clarity and accuracy.

Part (a): Square of an Even Integer Divisible by 4

Given:
We 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: $n = 2k \quad \text{for some integer } k.$

Now, squaring $n$: $n^2 = (2k)^2 = 4k^2.$

Clearly, $n^2$ is divisible by 4, since $n^2 = 4k^2$. Therefore, when the square of an even integer is divided by 4, it leaves a remainder of 0.

Correction Notes: The original solution contains a typo ("ni" instead of "n^2") and some unclear symbols in "ni = (27)3=4%3." The correct expression for squaring $n$ and showing divisibility by 4 is given above.

Part (b): Square of an Odd Positive Integer Divided by 4

Given:
We are asked to show that the square of an odd positive integer leaves a remainder of 1 when divided by 4.

Solution:
Let $n$ be an odd integer. Thus, $n$ can be written as: $n = 2k + 1 \quad \text{for some integer } k.$

Now, squaring $n$: $n^2 = (2k + 1)^2 = 4k^2 + 4k + 1 = 4(k^2 + k) + 1.$

So, $n^2$ is of the form $4m + 1$, where $m = k^2 + k$. This means when $n^2$ is divided by 4, the remainder is 1.

Correction Notes: The solution is mostly correct, but the original uses unclear notation, such as "4RP" and "4(k2+k)." These should be corrected to the proper algebraic expressions.

Part (c): Identisquares

Given:
We are asked to show that there are no "identisquares" 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 the 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:

$11^2 = 121$ $22^2 = 484$ $33^2 = 1089$ $44^2 = 1936$ $55^2 = 3025$ $66^2 = 4356$ $77^2 = 5929$ $88^2 = 7744$ $99^2 = 9801$

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

Correction Notes: The solution is correct, but it is important to note that the squares of two-digit numbers indeed do not result in identisquares. The examples given adequately demonstrate this.

Summary

• Part (a): The even integer case was mostly correct but had some unclear notation.
• Part (b): The odd integer case was correct, but again, some notation needed clarification.
• Part (c): The identisquares case was correctly solved, and the reasoning was sound.

Would you like more details or have any questions on this topic?

Here are 5 related questions to consider:

1. How can you determine if a number is divisible by a given number?
2. What are the general properties of squares of integers?
3. How does modular arithmetic help in solving divisibility problems?
4. Are there other examples of numbers that have unique properties when squared?
5. How does the concept of perfect squares relate to number theory?

Tip: When solving problems with modular arithmetic, double-check each step to ensure accuracy, especially when simplifying expressions.