If x and y are integer, what is the remainder when x^2 + y^2 is divided by 5? When x-y...

Understanding the Question

We need to find the remainder when \(\mathrm{x}^2 + \mathrm{y}^2\) is divided by 5, where x and y are integers.

Key Insight: The Pattern of Squares

Here's what makes this problem manageable: when we square any integer and divide by 5, we can only get three possible remainders: 0, 1, or 4. Let's see why:

• Numbers ending in 0 or 5: their squares end in 0 → remainder 0
• Numbers ending in 1, 4, 6, or 9: their squares end in 1 or 6 → remainder 1
• Numbers ending in 2, 3, 7, or 8: their squares end in 4 or 9 → remainder 4

This dramatically limits our possibilities! For the statements to be sufficient, they must force \(\mathrm{x}^2 + \mathrm{y}^2\) to have exactly ONE possible remainder when divided by 5.

Analyzing Statement 1

Statement 1 tells us: When x - y is divided by 5, the remainder is 1.

In other words: \(\mathrm{x} - \mathrm{y} \equiv 1 \pmod{5}\)

Let's test strategic examples to see if this pins down a unique remainder for \(\mathrm{x}^2 + \mathrm{y}^2\):

Example 1: Let x = 6 and y = 5

• Check: x - y = 1, which has remainder 1 when divided by 5 ✓
• Calculate: \(\mathrm{x}^2 + \mathrm{y}^2 = 36 + 25 = 61\)
• \(61 \div 5 = 12\) remainder \(1\)

Example 2: Let x = 7 and y = 6

• Check: x - y = 1, which has remainder 1 when divided by 5 ✓
• Calculate: \(\mathrm{x}^2 + \mathrm{y}^2 = 49 + 36 = 85\)
• \(85 \div 5 = 17\) remainder \(0\)

We found two different remainders (1 and 0) from pairs that both satisfy Statement 1.

[STOP - Not Sufficient!]

Statement 1 is NOT sufficient.

This eliminates choices A and D.

Analyzing Statement 2

Now we forget Statement 1 completely and analyze Statement 2 independently.

Statement 2 tells us: When x + y is divided by 5, the remainder is 2.

In other words: \(\mathrm{x} + \mathrm{y} \equiv 2 \pmod{5}\)

Let's test different scenarios:

Example 1: Let x = 1 and y = 1

• Check: x + y = 2, which has remainder 2 when divided by 5 ✓
• Calculate: \(\mathrm{x}^2 + \mathrm{y}^2 = 1 + 1 = 2\)
• \(2 \div 5 = 0\) remainder \(2\)

Example 2: Let x = 2 and y = 0

• Check: x + y = 2, which has remainder 2 when divided by 5 ✓
• Calculate: \(\mathrm{x}^2 + \mathrm{y}^2 = 4 + 0 = 4\)
• \(4 \div 5 = 0\) remainder \(4\)

Again, we get different remainders (2 and 4) from pairs that satisfy Statement 2.

[STOP - Not Sufficient!]

Statement 2 is NOT sufficient.

This eliminates choice B (and confirms D is already eliminated).

Combining Both Statements

Now we use BOTH conditions together:

• \(\mathrm{x} - \mathrm{y} \equiv 1 \pmod{5}\)
• \(\mathrm{x} + \mathrm{y} \equiv 2 \pmod{5}\)

Here's the key insight: these two conditions form a system that uniquely determines what remainders x and y must have when divided by 5.

Why this matters:

• If we add the conditions: \((\mathrm{x} - \mathrm{y}) + (\mathrm{x} + \mathrm{y}) = 2\mathrm{x} \equiv 1 + 2 \equiv 3 \pmod{5}\)
• If we subtract them: \((\mathrm{x} + \mathrm{y}) - (\mathrm{x} - \mathrm{y}) = 2\mathrm{y} \equiv 2 - 1 \equiv 1 \pmod{5}\)

Since we're working modulo 5 (where 5 is odd), each of these equations has exactly one solution:

• From \(2\mathrm{x} \equiv 3 \pmod{5}\), we can determine x (mod 5) uniquely
• From \(2\mathrm{y} \equiv 1 \pmod{5}\), we can determine y (mod 5) uniquely

Since x and y are uniquely determined modulo 5, and since \(\mathrm{x}^2\) and \(\mathrm{y}^2\) only depend on the remainders of x and y when divided by 5, the value of \(\mathrm{x}^2 + \mathrm{y}^2\) modulo 5 is also uniquely determined.

[STOP - Sufficient!]

Both statements together are sufficient.

This eliminates choice E.

The Answer: C

Both statements together uniquely determine the remainder, but neither statement alone is sufficient.

Answer Choice C: "Both statements together are sufficient, but neither statement alone is sufficient."