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Sum of squares

Show that if \(a \geq b \geq 2 \) are integers such that \(2ab\) is a perfect square, then \( a^2 + b^2 \) is not a prime.

Note by Chung Kevin
3 years, 7 months ago

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Let \(2ab=c^2. Then (a+b)^2 - 2ab = a^2 + b^2 =(a+b)^2 - c^2 = (a+b+c)(a+b-c)\), which is not prime because a+b-c cannot be equal to 1(a and b are bigger or equal to 2). Bogdan Simeonov · 3 years, 7 months ago

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@Bogdan Simeonov \(a + b - c\) is not equal to 1 or -1, because we will get \(a + b \pm 1 = c\) and then by squaring both of the sides we will get \(a^2 + b^2 + 1 + 2ab \pm2a \pm 2b = 2ab\), which is impossible. Bogdan Simeonov · 3 years, 7 months ago

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