Diophantine question: find two positive numbers a and b such that $a\cdot b+a=c^3$ and $a\cdot b+b=d^3$
I solved this equation , the solution are rational; $\frac{112}{13}$ and $\frac{27}{169}$. Can this problem have more solutions?
Solution: We set parametric $a=t^3x$ and $b=x^2-1$.We have:
$a\cdot b=t^3x(x^2-1)=(tx)^3-t^3x$
Such that:
$a\cdot b+a=(tx)^3-t^3x+t^3x=(tx)^3=c^3$
for b we must have:
$a\cdot b+b=(tx)^3-t^3x+x^2-1=d^3$
Now let $(tx)^3-t^3x+x^2-1=(tx-1)^3$, then we have:
$(tx-1)^3=(tx)^3-3t^2x^2+3tx-1=(tx)^3+ x^2-t^3x-1$
Which gives:
$-3t^2x^2+3tx=x^2-t^3x$
$x[(t^3+3t)-(3t^2+1)x]=0$
$x=\frac{t^3+3t}{3t^2+1}$
This is parametric solution of x in terms of t. Hence this question can have infinite solutions.With $t=2$ ,I found $x=\frac{14}{13}$ which gives $a=\frac{112}{13}$ and $b=\frac{27}{169}$.
According to sirous's method, we can get another parametric solution as follows.
From first equation, let $a=ct$ and $b=c^2/t-1$.
Substitute it to second equation, we obtain
$$c^3-ct+c^2/t-1 = d^3\tag{1}$$
Substitute $d=c-1$ to equation $(1)$, then we get $c = \frac{t(3+t)}{1+3t}$.
$$(a,b)=\left(\frac{t^2(3+t)}{1+3t}, \frac{(t-1)^3}{(1+3t)^2}\right)$$ $t$ is arbitrary.
If $t>1$ then $a$ and $b$ are positive.
Example:
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$$ \left\{ \begin{array}{c} ab+a = c^3 \\ ab+b = d^3 \\ \end{array} \right. $$
We can get a parametric solution as follows (partial solution).
From first equation, let $a=c$ and $b=c^2-1$.
Substitute it to second equation, we obtain
$$c^3 + c^2 - c -1 = d^3\tag{1}$$
Since equation $(1)$ has a solution $(c,d)=(1,0)$, then we can get a parametric solution below.
$$(a,b)=\left(\frac{-(k+1)(k^2-k+1)}{(k-1)(k^2+k+1)}, \frac{4k^3}{(k-1)^2(k^2+k+1)^2}\right)$$ $k$ is arbitrary.
If $0<k<1$ then $a$ and $b$ are positive.
Example: