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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). · 3 years, 7 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. · 3 years, 7 months ago