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The Equation $A^2 - 4B = 3$ Has No Integer Solution

Alexander Ross • • 2 min read
The Equation $A^2 - 4B = 3$ Has No Integer Solution
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Prove that: $$a^2 - 4b \neq 3$$ for all integers $a$ and $b$.

I'm not sure how to find a way to prove this statement. Some help would be appreciated.

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2 Answers

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Assume that it is true. So $a^2-4b=3$.

Case I :When $a$ is even. Then $a=2k$ where $k$ is an integer. Thus we have, $4(k^2-b)=3$. LHS is even and RHS is odd. Contradiction.

CaseII: When $a$ is odd. Then $a=2k+1$ where $k$ is an integer. Then, $(2k+1)^2-4b=3$, hence we have $2(k^2+k-b)=1$. Again LHS even, RHS odd. Contradiction.

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Take $\mod4$

LHS $\equiv 0,1 \pmod 4$ but RHS $\equiv 3$

So no solutions

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Alexander Ross
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Alexander Ross

Alexander Ross has covered the video game industry for a decade, writing deep dives on game design, esports tournaments, VR developments, and gaming culture.