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).
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Bogdan Simeonov
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3 years, 3 months ago

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@Bogdan Simeonov
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\(a + b - c\) 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.
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Bogdan Simeonov
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3 years, 3 months ago

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TopNewestLet \(2ab=c^2. Then (a+b)^2 - 2ab = a^2 + b^2 =(a+b)^2 - c^2 = (a+b+c)(a+b-c)\), 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, 3 months ago

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– Bogdan Simeonov · 3 years, 3 months ago

\(a + b - c\) 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.Log in to reply