Solve the diophantine equation $b^2ac + b(c-a^2-1) +(7c -a) = 0$

Question

Question:

Solve the following diophantine equation:
$b^2ac + b(c-a^2-1) +(7c -a) = 0$ where $a,b,c∈\mathbb{N}$

My approach:

Find roots of this quadratic equation and then proceed to make the discriminant real.
$(c-a^2-1)^2 \ge 4ac(7c-a)$ $\qquad(1)$

Similarly if we rearrange as quadratic of $a$:
$a^2(b) +a(1-b^2c) + b- bc- 7c = 0$
$(1-b^2c)^2 \ge 4b(b- bc- 7c)$ $\qquad(2)$

Other thought process I have is , since coefficient of $a^2$ and $b^2$ are of opposite sign we might be able to express the equation as difference of two whole squares like Pell's equation.

Answer 1

I will frequently use if $m,n\in \mathbb{N}$ and $m\mid n$ then $m\leq n$.

So we have $$ab^2+b+7\mid a^2b+a+b$$ hence $$ab^2+b+7\mid (ab^2+b+7)a- (a^2b+a+b)b = 7a-b^2$$

• If $7a>b^2$ then there is positive $x$ such that $$x(ab^2+b+7)=7a-b^2\implies a(xb^2-7)= -b^2-xb-7x<0 \implies xb^2<7$$

So if $b>2$ there is no solution. If $b=2$ we have $x=1$ and thus $3a=13$ so no solution. If $b=1$ then $a+8\mid 7a-1$ so $$a+8\mid (a+8)7-(7a-1)=57 \implies a\in\{ 11,49\}$$

• If $7a<b^2$ then $$ab^2+b+7\mid b^2-7a\implies b^2(a-1)+b+7(1+a)\leq 0$$ This quadratic inequality has a solution iff $$1-28(a^2-1) \geq0$$ But now we have $28a^2\leq 29\implies a=1$. This yields $$b^2+b+7\mid 2b+1\implies b^2-b+6\leq 0$$ which is not true.

• So we are left with $7a=b^2$ and this case is not difficult to handle...