a^2 - b = 133 (1)
b^2 - a = 133 (2)
The original problem, about which there are several
Youtube videos,
has the additional condition a ≠ b. But I did not see this in the banner
pictures. So I found more solutions.
All the Youtube solutions I watched subtracted the two equations (1) and (2) and transformed the result (more or less) like this:
a^2 - b - b^2 + a = 0
a^2 - b^2 = - (a - b)
(a + b) (a - b) = - (a - b)
Divide both sides by a - b, which is possible since a ≠ b, and isolate b:
a + b = -1
b = -1 - a
Use this to eliminate b in equation (1):
a^2 - (-1 - a) = 133
a^2 + a - 132 = 0
From this quadratic equation two solutions for a (and then also for b) can be computed easily.
(See below.)
The core trick above was to rewrite a^2 - b^2 as (a + b)(a - b),
and to cancel the factor a - b with another occurrence of that expression
that happens to appear "magically".
I did not see this possibility and therefore used a different approach.
Isolate b in equation (1) and substitute the expression for b in equation (2).
Then transform this into a quartic equation with the single variable a:
b = a^2 - 133
(a^2 - 133)^2 - a = 133
a^4 - 266 a^2 + 17689 - a - 133 = 0
a^4 - 266 a^2 - a + 17556 = 0
Let's call the LHS polynomial P4.
There is a formula/algorithm for solving quartic equations, but I don't know it.
Due to the symmetry of the problem we can hope for solutions with a = b.
This assumption simplifies both of the original equations (1) and (2) to
a^2 - a - 133 = 0
where we call the LHS polynomial P2.
The equation has solutions a = (1 ± sqrt(533))/2.
In each case b is the same as a.
But we expect two more solutions for the quartic equation (actually the ones with a ≠ b). For this we could
divide P4 by (a - (1 + sqrt(533))/2) and then by (a - (1 - sqrt(533))/2), which
results in a polynomial of degree 2. But it is easier to directly divide P4
by P2 (which is just the product of these two divisors).
(a^4 - 266 a^2 - a + 17556) / (a^2 - a - 133) = a^2 + a - 132
a^4 - a^3 - 133 a^2
-------------------------------
a^3 - 133 a^2 - a + 17556
a^3 - a^2 - 133 a
-------------------------
- 132 a^2 + 132 a + 17556
- 132 a^2 + 132 a + 17556
-------------------------
0
As expected, the division leaves no rest and the quotient polynomial has degree 2. Setting that polynomial to zero yields the quadratic equation that was also derived in the "standard" solution above:
a^2 + a - 132 = 0
With the quadratic formula we get the two solutions
a = 11 and a = -12
For symmetry reasons in each case b must be the other one of these solutions.
(This is also straight-forward to compute if you do not believe
the symmetry argument.)