Does $4b^3pq=2a^2+b$ has a solution for non-zero integers $a,b$ and odd primes $p<q$?

Question

Just recently, I had an equation on my hands that I thought had no solutions: Why $4b^3p=2a^2+qb$ has no solutions for integers $a,b$ and two odd primes $p<q$?

I was on a wrong path and now have an equation $4b^3pq=2a^2+b$ that really do not brings me any non-zero integer solution $a,b$ where $2<p<q$ are odd primes. What I see is that $b$ must be even and when I substitute $b=2t$ I have to solve $0=16t^3pq-t-a^2$ for $t$ (quadratic equation) and analyze the discriminant?

Answer 1

The correct resulting equation is of substituting $b=2t$ is: $$16t^3pq=a^2+t$$

We transform our equation to: $$(16t^2pq-1)t = a^2$$

Let $\ell$ be any prime factor of $t$, and let $\ell^n$ be the highest power of $\ell$ that divides $\color{red}{a}$. Then $\ell^{2n}$ is the highest power of $\ell$ that divides $a^2$; now $\ell$ does not divide $16t^2pq - 1$ at all, so $\ell^{2n}$ must also be the highest power of $\ell$ that divides $t$.

Since $\ell$ is arbitrary, we conclude that $t$ is in fact a square, say $t = g^2$. Then our equation becomes: $$(16g^4pq-1)g^2 = a^2$$

which gives: $$16g^4pq-1 = \left(\frac ag\right)^2$$

which is impossible, since the squares mod $16$ are $0, 1, 4, 9$.

Answer 2

More generally, the equation $$ 2b^3c=2a^2+b \qquad\;\;\;\; $$ has no integer solutions with $c > 0$ and $a,b\ne 0$.

Suppose otherwise. \begin{align*} \text{Then}\;\;& 2b^3c=2a^2+b \qquad\qquad\qquad\; \\[4pt] \implies\;& 2b^3c-b=2a^2 \\[4pt] \implies\;& b(2b^2c-1)=2a^2 \\[4pt] \implies\;& b(2b^2c-1) > 0 \\[4pt] \implies\;& b > 0\;\;\text{(since $2b^2c-1 > 0)$} \end{align*} Next let $d=\gcd(a,b)$.

Then $$ \left\lbrace \begin{align*} a&=da_1\\[4pt] b&=db_1\\[4pt] \end{align*} \right. \qquad\qquad\qquad\;\;\; $$ for some integers $a_1,b_1$ with $b_1 > 0$ and $\gcd(a_1,b_1)=1$. \begin{align*} \text{Then}\;\;& 2b^3c=2a^2+b \qquad\qquad\qquad\;\;\;\; \\[4pt] \implies\;& 2(db_1)^3c=2(da_1)^2+db_1 \\[4pt] \implies\;& 2d^3b_1^3c=2d^2a_1^2+db_1 \\[4pt] \implies\;& 2d^2b_1^3c=2da_1^2+b_1 \\[4pt] \implies\;& 2d{\,\mid\,}b_1 \\[4pt] \end{align*} hence $b_1=2db_2$ for some positive integer $b_2$.

Then since $\gcd(a_1,b_1)=1$ and $b_2{\,\mid\,}b_1$, it follows that $\gcd(a_1,b_2)=1$. \begin{align*} \text{Then}\;\;& 2d^2b_1^3c=2da_1^2+b_1 \qquad\qquad\qquad \\[4pt] \implies\;& 2d^2(2db_2)^3c=2da_1^2+2db_2 \\[4pt] \implies\;& 16d^5b_2^3c=2da_1^2+2db_2 \\[4pt] \implies\;& 8d^4b_2^3c=a_1^2+b_2 \\[4pt] \implies\;& b_2{\,\mid\,}a_1^2\ \\[4pt] \implies\;& b_2=1\;\;\text{(since $\gcd(a_1,b_2)=1$)} \\[4pt] \implies\;& 8d^4c=a_1^2+1 \\[4pt] \implies\;& a_1^2\equiv -1\;(\text{mod}\;4) \\[4pt] \end{align*} which is not possible.