It is clear that we can generate infinitely many Pythagorean triples (x, y, z) b

Question

It is clear that we can generate infinitely many Pythagorean triples (x, y, z) but not clear (even from Euclid's formulas) whether there are any significant constraints on their members x, y, and z. For example, can we have x and y odd and z even? This question can be answered by considering remainders on division by 4. Show that the square of an odd integer 2n + 1 leaves remainder 1 on division 4. What is the remainder when an even square is divided by 4? Deduce from Exercises 1.8.4 and 1.8.5 that the sum of odd squares is never a square.

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Dear Student Thank you for using Chegg !! Given an odd integer (2n + 1) To prove (2n + 1)^2 / 4 leaves a remainder of 1 Solution Squaring (2n+ 1) we get (2n + 1)^2 = (2n)^2 + 2. 2n + 1 (Using (a+b)^2 = a^2 + b^2 + 2ab ) Gives (2n + 1)^2 = 4n^2 + 4n + 1 Now dividing the equation by 4 we get (2n + 1)^2 / 4 = (4n^2 + 4n + 1 )/ 4 = n^2 + n + (1/4) Hence clearly only the 1 in the expansion of (2n + 1)^2 is left undivided when divided by 4 which implies that 1 is the remainder Hecne proved (2n + 1)^2 / 4 leaves a remainder of 1 Solution

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