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I try to find integers $a$ and $b$ such that $a^2+b^2 = 10^{100}+3$, I try some number without any result

Bernard
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Fouad El
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    Note that $10^{100}+3\equiv3\pmod4$. Also note that a perfect square $\pmod4$ is either $0$ or $1$. What are the possibilities for $a^2+b^2\pmod4$? – Aiden Chow Feb 11 '21 at 21:57

1 Answers1

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hint

$a^2+b^2 $ is odd, so $ a$ and $ b $ have not the same parity.

Assume $$a=2A \; \text{ and } \; b=2B+1$$

thus

$$4A^2+4B^2+4B=10^{100}+2$$ or

$$2A^2+2B^2+2B=5.10^{99}+1$$ the LHS is even, while the RHS is not.

hamam_Abdallah
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