Diophantine equation from the Latvian Baltic Way team selection competition 2019

Question

So here is the problem statement:

Find all integer triples $(a, b, c)$ such that

$(a-b)^3(a+b)^2 = c^2 + 2(a-b) + 1$

The only things I have so far figured out is that (-1, 0, 0) and (0, 1, 0) are solution, gcd((a-b), c) = 1 and that c must be even.

Any ideas? Maybe even a general solution to the problem?

Answer 1

Consider the equation already found:- $$(a-b)\left((a^2-b^2)^2-2\right)=c^2+1,$$ where $c^2+1$ is odd or singly even.

If $a-b$ is even then the LHS would be divisible by 4. Therefore $a-b$ is odd and $c$ is even.

Then $(a^2-b^2)^2-2$ is of the form $4k-1$ and it is therefore either $-1$ or has a prime divisor of the form $4k-1$. However, $c^2+1$ cannot have a prime factor of the form $4k-1$ and so $$(a^2-b^2)^2-2=-1.$$

Therefore $a^2-b^2$ is $-1$ or $1$ and then one of $a$ and $b$ is $0$. We also have $b-a=c^2+1$ and so the solutions for $(a,b,c)$ are $(0,1,0)$ or $(-1,0,0).$

Answer 2

Write as $(a-b)((a^2 - b^2)^2 - 2) = c^2 + 1$. Use that there is no prime dividing $c^2 + 1$ of the form $4k+3$. When are the multipliers of the LHS both congruent to 1 mod 4 or are such that the product is 1 in modulus?

Answer 3

Above equation shown below:

$(a-b)^3(a+b)^2 = (c^2 + 2(a-b) + 1)$ ----$(1)$

Equation $(1)$ has parameteric solution given below:

$(a,b,c)=[(2m^2+2),(2m^2+1),(2m)]$

For, $m=7$ we get:

$(a,b,c)=(100,99,14)$

Answer 4

Equation, $(a-b)^3(a+b)^2 = (c^2 + 2(a-b) + 1)$

Thanks @Piquito, for reviewing my previous answer.

By mistake I solved "OP' equation as a quartic rather than a quantic.

If we put the condition, $(a=b+c)$ then we get:

$c^3(2b+c)^2=(c+1)^2$

Which has solution at, $(b,c)=[(1/2),(1)]$

And so we get, $(a,b,c)=[(3/2),(1/2),(1)]$

Answer 5

COMMENT.- (another way) If $a,b,c\in\mathbb Z$ then $a+b,a-b\in\mathbb Z$ so we can study the diophantine equation $$d^3s^2=c^2+2d+1$$ Taking now $d$ as a parameter, $x=s$ and $y=c$ we have a family of conics $\Gamma_d$ defined by $$\Gamma_d:\quad d^3x^2-y^2-(2d+1)=0$$ $d$ cannot be zero clearly in the original equation.

Whith $d=a-b\gt0$ we have an hyperbola parameterized by $$x=\sqrt{\dfrac{2d+1}{d^3}}\sec(t)\\y=\sqrt{2d+1}\tan(t)$$ Whith $d=a-b\lt0$ we have an ellipse parameterized by $$x=\sqrt{\dfrac{2d+1}{d^3}}\cos(t)\\y=\sqrt{2|d|+1}\sin(t)$$

It appears as highly probable that there are not non-trivial solutions.