Find three rational numbers $a,b,c$ s.t. $c^2-b^2=b^2-a^2=1111$.

Question

Find three rational numbers $\,a,b,c\,$ s.t. $\,c^2-b^2=b^2-a^2=1111\,$.

\begin{alignat*}{2} \left(\dfrac{10199}{60}\right)^2&-&\left(\dfrac{10001}{60}\right)^2&=1111\\ \left(\dfrac{10001}{60}\right)^2&-&\left(\dfrac{9799}{60}\right)^2&=1111\\ \end{alignat*}

What will be the next set of solutions?

Answer 1

\begin{align*} \begin{cases} c^2-b^2=N\\ b^2-a^2=N \end{cases}\quad \xrightarrow[\phantom{\hspace{3cm}}]{\begin{split} \displaystyle{v=\dfrac{a^4+b^4}{4abc}} \end{split}}\quad \begin{cases} w^2-v^2=N\\ v^2-u^2=N \end{cases} \end{align*} We can find that \begin{alignat*}{2} \left(\dfrac{10788083791879201}{119939994000120}\right)^2&-&\left(\dfrac{10019997400200001}{119939994000120}\right)^2&=1111\\ \left(\dfrac{10019997400200001}{119939994000120}\right)^2&-&\left(\dfrac{9187923807880801}{119939994000120}\right)^2&=1111\\ \end{alignat*}

Cf. Recreations in the Theory of Numbers: The Queen of Mathematics Entertains

Answer 2

$$a^2+c^2=2b^2$$

If $a=2pq,c=p^2-q^2,2b^2=p^2+q^2$

WLOG $p=m+n,q=m-n$

$\implies b^2=m^2+n^2$

and $c^2=?$

$$1111=c^2-b^2=(4mn)^2-(m^2+n^2)^2=(4mn+m^2-n^2)(4mn-(m^2-n^2))$$

Now $1111=1\cdot1111=11\cdot101$

Answer 3

$$c^2-b^2=1111$$ $$b^2-a^2=1111$$ From first equation, we get the solution $$b = 1/2(11t^2-101)/t, c = 1/2 (11t^2+101)/t.$$

t is arbitrary.
Substitute $b = 1/2(11t^2-101)/t$ to second equation, we get $$a^2 = 1/4(121t^4-6666t^2+10201)/(t^2).$$ Let $v=2ta$, we consider the quartic equation below.
$$v^2 = 121t^4-6666t^2+10201$$ This quartic equation can be transformed to the elliptic curve $$Y^2=X^3 -1234321X$$
This elliptic curve has rank 1 with generator $P(X,Y)=(-1111/100, -3702963/1000)$.
Thus, this elliptic curve has infinitely many rational points.
$P: (a,b,c)=(\frac{9799}{60}, \frac{10001}{60},\frac{10199}{60})$
$2P: (a,b,c)=(\frac{9187923807880801}{119939994000120}, \frac{10019997400200001}{119939994000120}, \frac{10788083791879201}{119939994000120})$
$3P: (a,b,c)=(\frac{814164545232834936227590554656568199}{17956778167727434877277815956800180}, \frac{1010495562483810981102483955601050001}{17956778167727434877277815956800180}, \frac{1174452492391564623814785274944571799}{17956778167727434877277815956800180})$