Wading Herons

The sides of all isosceles Heronian triangles, for any integers $x,y$ can be expressed as

$A=B={ x }^{ 2 }+{ y }^{ 2 }$
$C=2({ x }^{ 2 }-{ y }^{ 2 })$

Hence, if $A=B=n$ and $C=n+1$ , we end up with the Pell Equation

${ x }^{ 2 }-3{ y }^{ 2 }=1$

The first integer solution $\left\{ x,y \right\}$ . is $\left\{ 2,1 \right\}$ . The $y$ of the next can be found by this recursion formula

${ y }_{ i+1 }=2{ y }_{ i }+{ x }_{ i }$

so that we have the sequence of solutions

$\left\{ 2,1 \right\} ,\left\{ 7,4 \right\} ,\left\{ 26,15 \right\} ,\left\{ 97,56 \right\} ,\left\{ 362,209 \right\} ,\left\{ 1351,780 \right\} ,\left\{ 5041,2911 \right\} ,...$

The $4$ th smallest Heronian triangle meeting the conditions would be based on $\left\{ 97,56 \right\}$

$n={ 97 }^{ 2 }+{ 56 }^{ 2 }=12545$
$n+1=2 ( { 97 }^{ 2 }-{ 56 }^{ 2 } ) =12546$

so that the area works out to $68149872$ and the rest is calculation work

The hint really helped me..nice question