How can I show that for every pair of non-negative integers $(b,c)$, the expression
$$4bc + 3b + 3c + 2$$
is not a perfect square?
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Hint:
Multiply by $4$ and you get $(4b+3)(4c+3)-1$.
If this is a square then $(4b+3)(4c+3)$ can be written as the sum of two squares.
Thomas Andrews
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Why would that be the case then? – Måns Nilsson Jul 13 '17 at 15:44
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Why would what even be the case? There are a lot of claims up there. Help me help you. – Thomas Andrews Jul 13 '17 at 15:44
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How do you deduce what you're saying in your last sentence? – Måns Nilsson Jul 13 '17 at 15:45
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2If $(4b+3)(4c+3)-1$ is a perfect square, then $(4b+3)(4c+3)$ is one plus a perfect square. – Thomas Andrews Jul 13 '17 at 15:46
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To add to the answer by Thomas, with some integer $x,$ $1 + x^2$ is never divisible by any (positive) prime $q \equiv 3 \pmod 4.$ The quick reason for this is that $-1$ is not a quadratic residue $\pmod q.$ Writing the Legendre symbol $(-1|q) = -1.$
I wrote a short proof of the general fact at Prime divisors of $k^2+(k+1)^2$
Will Jagy
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