AoPS: Lesson 5, Class Extensions' Solutions

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Problem 1: Compute the following:

(a) \((1 + 2i)(1 − 2i)\)

Use the difference of squares formula:

\[ (1 + 2i)(1 − 2i) = 1^2 − (2i)^2 = 1 − 4i^2 = 1 − 4(−1) = 1 + 4 = 5 \]

(b) \((3 + i)(3 − i)\)

Again, use the difference of squares formula:

\[ (3 + i)(3 − i) = 3^2 − (i)^2 = 9 − (−1) = 9 + 1 = 10 \]

(c) \((2 + 3i)(2 − 3i)\)

Apply the difference of squares:

\[ (2 + 3i)(2 − 3i) = 2^2 − (3i)^2 = 4 − 9(−1) = 4 + 9 = 13 \]

Observation: In each case, the product of conjugates results in a real number. Specifically, the sum of the squares of the real and imaginary parts of the complex numbers yields the product.

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Problem 2:

(a) Write \( 61 \) as a sum of two perfect squares.

We can express \( 61 \) as:

\[ 61 = 5^2 + 6^2 \]

(b) Use this to write \( 61 \) as \( (a + bi)(a − bi) \) where \( a \) and \( b \) are integers.

From part (a), we can write:

\[ 61 = (5 + 6i)(5 − 6i) \]

Thus, \( a = 5 \) and \( b = 6 \).

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Problem 3:

(a) Compute the product \( (1 + 2i)(4 − i) \).

Use the distributive property:

\[ (1 + 2i)(4 − i) = 1(4 − i) + 2i(4 − i) = 4 − i + 8i − 2i^2 = 4 + 7i − 2(−1) = 4 + 7i + 2 = 6 + 7i \]

(b) Show that the product \( (1 + 2i)(4 − i)(1 − 2i)(4 + i) \) is a sum of two perfect squares.

First, simplify the two pairs:

\[ (1 + 2i)(1 − 2i) = 1^2 − (2i)^2 = 1 + 4 = 5 \]

\[ (4 − i)(4 + i) = 4^2 − i^2 = 16 + 1 = 17 \]

Now multiply the results:

\[ 5 \times 17 = 85 \]

Since \( 85 = 9^2 + 2^2 \), this is the sum of two perfect squares.

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Problem 4:

Show that the product \( (1 + 2i)(4 − i)(1 − 2i)(4 + i) \) is a product of two numbers, each of which is the sum of two perfect squares.

We already computed \( (1 + 2i)(1 − 2i) = 5 \) and \( (4 − i)(4 + i) = 17 \), which are both sums of two perfect squares:

\[ 5 = 2^2 + 1^2, \quad 17 = 4^2 + 1^2 \]

Thus, the product \( (1 + 2i)(4 − i)(1 − 2i)(4 + i) \) can be expressed as the product of two sums of squares.

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Problem 5:

(a) Write \( 13 \) as the sum of two perfect squares.

We can express \( 13 \) as:

\[ 13 = 3^2 + 2^2 \]

(b) Write \( 82 \) as the sum of two perfect squares.

We can express \( 82 \) as:

\[ 82 = 9^2 + 1^2 \]

(c) Write \( 13 \times 82 \) as the sum of two perfect squares.

We use the identity for the product of two sums of squares:

\[ (3^2 + 2^2)(9^2 + 1^2) = (3 \times 9 − 2 \times 1)^2 + (3 \times 1 + 2 \times 9)^2 = 25^2 + 21^2 = 625 + 441 = 1066 \]

Thus, \( 13 \times 82 = 25^2 + 21^2 \).

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Problem 6:

Show that \( (a^2 + b^2)(c^2 + d^2) \) is the sum of two perfect squares.

We use the identity for the product of two sums of squares:

\[ (a^2 + b^2)(c^2 + d^2) = (ac − bd)^2 + (ad + bc)^2 \]

Thus, the product of two sums of squares is itself the sum of two squares.

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Problem 7:

(a) Find a way of writing any square of a sum of squares, \( (m^2 + n^2)^2 \), with \( m > n \), as a sum of two positive squares.

We can expand:

\[ (m^2 + n^2)^2 = m^4 + 2m^2n^2 + n^4 = (m^2 − n^2)^2 + (2mn)^2 \]

Thus, \( (m^2 + n^2)^2 = (m^2 − n^2)^2 + (2mn)^2 \), which is a sum of two positive squares.

(b) Find a right triangle with integer sides and a hypotenuse of length \( 13 \times 82 \).

From Problem 5(c), we know \( 13 \times 82 = 25^2 + 21^2 \). Thus, the right triangle has sides \( 21 \) and \( 25 \) with hypotenuse \( \sqrt{1066} \).