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Hill

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- Homework Statement
- "Here is a basic fact about integers that has many uses in number theory: If two integers can be expressed as the sum of two squares, then so can their product. With the understanding that each symbol denotes an integer, this says that if ##M = a^2 + b^2## and ##N = c^2 + d^2##, then ##MN = p^2 + q^2##. Prove this result by considering ##|(a + ib)(c + id)|^2##."

- Relevant Equations
- ##|x+iy|^2 = x^2 + y^2##

Firstly, the exercise itself is not difficult:

On one hand, $$|(a + ib)(c + id)|^2 = |a + ib|^2|c + id|^2 = (a^2 + b^2) (c^2 + d^2) = MN.$$

On the other hand, ##(a + ib)(c + id) = p+ iq## for some integers p and q, and so $$|(a + ib)(c + id)|^2 = |p + iq|^2 = p^2 + q^2.$$

Thus, ##MN = p^2 + q^2.##

However, it can be done quite straightforwardly, without considering the complex numbers, e.g.,

$$MN = (a^2 + b^2) (c^2 + d^2) = a^2 c^2 + a^2 d^2 + b^2 c^2 + b^2 d^2 =$$$$a^2 c^2 + a^2 d^2 + b^2 c^2 + b^2 d^2 + 2abcd - 2abcd = $$$$(a^2 c^2 + 2abcd + b^2 d^2) + (a^2 d^2 - 2abcd + b^2 c^2) =$$$$(ac + bd)^2 + (ad - bc)^2 = p^2 + q^2.$$

I don't see an advantage of considering the complex numbers in this case. What am I missing?

On one hand, $$|(a + ib)(c + id)|^2 = |a + ib|^2|c + id|^2 = (a^2 + b^2) (c^2 + d^2) = MN.$$

On the other hand, ##(a + ib)(c + id) = p+ iq## for some integers p and q, and so $$|(a + ib)(c + id)|^2 = |p + iq|^2 = p^2 + q^2.$$

Thus, ##MN = p^2 + q^2.##

However, it can be done quite straightforwardly, without considering the complex numbers, e.g.,

$$MN = (a^2 + b^2) (c^2 + d^2) = a^2 c^2 + a^2 d^2 + b^2 c^2 + b^2 d^2 =$$$$a^2 c^2 + a^2 d^2 + b^2 c^2 + b^2 d^2 + 2abcd - 2abcd = $$$$(a^2 c^2 + 2abcd + b^2 d^2) + (a^2 d^2 - 2abcd + b^2 c^2) =$$$$(ac + bd)^2 + (ad - bc)^2 = p^2 + q^2.$$

I don't see an advantage of considering the complex numbers in this case. What am I missing?

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