Find this following diophantine equation integer solution $$a^3+b^3=(2ab+1)^2$$
I think this equation only have two following solution $$(a,b)=(1,0),(0,1)$$ maybe this equation have no other solution? because can see wolframalpha.com
I think following idea is usefull $$a^3+b^3=(a+b)(a^2-ab+b^2)=(2ab+1)^2$$ since $(a+b,a^2-ab+b^2)=1$ so $a+b,a^2-ab+b^2$ are square numbers?
On
First look for the solutions of the equation $z^2=a^3+b^3$ which are given for instance in http://www.cecm.sfu.ca/~nbruin/thesis.pdf, lemma 3.2.6. Then for each family (give values to $s,t$), in order to find the possibles $a,b$ and check if $2ab+1=z.$
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Write $s = a + b$, $t = ab$. Since your equation is symmetric in $a$ and $b$ it can be written as an equation in $s$ and $t$: a solution $a,b$ of $$a^3+b^3=(2ab+1)^2$$ gives a solution $s,t$ of $$4t^2 + 3st + 4t = s^3 - 1.$$
I don't know if there is an easy way to solve this by hand, but since it defines an elliptic curve, many algorithms are available, and it turns out that the only solution is $(s,t) = (1,0)$, which corresponds to the solutions you found.
Let's rewrite equation in $s,t$ doetoe (edit: he deleted his answer, look at the end of mine) got as follows: $4t^2+(3s+4)t+(1-s^3)=0$, and try to solve it for $t$. Because $t$ is integer, we must get that discriminant is a perfect square: $\Delta=(3s+4)^2-4\cdot 4(1-s^3)=9s^2+24s+16-16+16s^3=s(16s^2+9s+24)$. It's easy to see that, assuming $a,b$ satisfy OP's equation, that $2\nmid s$, and we can check that $3\nmid s$: if $3\mid a,b$ then $3\mid a^3+b^3$ but $3\nmid (2ab+1)$, and if $a\equiv 1,b\equiv -1\mod 3$ then again $3\mid a^3+b^3$ but $3\nmid (2ab+1)$. So $s$ and $24$ are relatively prime, and thus $s$ and $16s^2+9s+24$ are, so, because these two multiply to a perfect square, both must be perfect squares. But $16s^2+9s+24=(4s+1)^2+s+23$. Now, for $s>\frac{20}{7}: (4s+1)^2<(4s+1)^2+s+23<(4s+2)^2$ and for $s<-23: (4s+1)^2>(4s+1)^2+s+23>(4s+2)^2$ (simple quadrtatic inequalities, can be checked with Wolfram Alpha). So, unless $-23\leq s\leq \frac{20}{7}$ $\Delta$ cannot be perfect square. One can check that in this interval $\Delta$ is square for $s=1$ or $s=-23$. For $s=1$ we get $t=0$ or $-\frac{7}{4}$. First leads to solutions you have found, second doesn't give integer solutions. $s=-23$ doesn't result in real solutions.
So only solutions are indeed $(0,1),(1,0)$.
EDIT: doetoe's answer has expressed OP's equality in terms of $s=a+b,t=ab$ and it wad form $4t^2+3ts+4t=s^3-1$, and equivalence can be easily verified by substitution and elementary transformations.