Find $a$, $b$ and $c$ that satisfy the following equations

Question

Find all integers $a$,$b$ and $c$ that satisfy the following equations

$a^2=bc+1\tag{i}$

$b^2=ac+1\tag{ii}$

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I tried solving came out with following results:

$(i)-(ii)$ gives $a+b+c=0$ which implies, $a^3+b^3+c^3=3abc$

Multiplying equation $(i)$ with $a$ and multiplying equation $(ii)$ with $b$ and then subtracting $(i)-(ii)$ we get $a^2+b^2+ab=1$ which can be written as $(a+b)^2-ab=1$ , which can be further written as $c^2-ab=1$.

Answer 1

We have $$a^2-b^2=c(b-a)\Rightarrow (a-b)(a+b+c)=0$$ $$\Rightarrow a=b\quad\text{or}\quad c=-a-b$$

Case 1 : $a=b$. We have $a^2=ac+1\Rightarrow c=a-\frac 1a$. So, since $\frac 1a$ has to be an integer, we have $a=\pm 1$. Thus, in this case, $(a,b,c)=(1,1,0),(-1,-1,0).$

Case 2 : $c=-a-b$. We have $a^2+b^2+ab=1,$ i.e. $\left(2a+b\right)^2+3b^2=4$. So, we have $$(2a+b,b^2)=(\pm 2,0),(\pm 1,1),$$ i.e. $$(a,b,c)=(1,0,-1),(-1,0,1),(0,1,-1),(-1,1,0),(1,-1,0),(0,-1,1).$$

Hence, the answer is $$(a,b,c)=(1,1,0),(-1,-1,0),(1,0,-1),(-1,0,1),(0,1,-1),(-1,1,0),(1,-1,0),(0,-1,1).$$

Answer 2

Note that you have $b^2-a^2=c(a-b).$ As $b^2-a^2=-(a-b)(a+b),$ then let's consider some cases :

• if $a=b$ : you have $a^2=ac+1$ which implies that $a\neq 0$ and $c=\frac{a^2-1}{a},$ who have to be an integer, so $a=\pm1$ and you can check these ($a=\pm 1, b=a, c=0$) are solutions.

• if $a\neq b$ : then you have $c=-(a+b)$ which implies $a^2=-b(a+b)+1$ and $b^2=-a(a+b)+1.$ These are polynomials, let's see the first one in $a$ and the second one in $b$. For the first one, you get that $\Delta=-3b^2+4$ which only allows $b=0,1,-1$ to get solutions. Roots would be integers : $a=0,-1,1.$ As it is the same for $a$ ($a=0,1,-1$ which implies $b=0,-1,1$), you have just to check what combinations $(a,b)\in\{0,-1,1\}^2$ works with $a\neq b$ and you will have your solutions.