YouTube’s algorithm also presented me with videos about equations that could be solved by noticing the symmetry in their form.
Example 1
x + y = 4 x^5 + y^5 = 464Here, we notice that if x = 2 + u then y = 2 - u and:
(2 + u)^5 + (2 - u)^5 = 464The odd powers of u will cancel out on the expansion of the term on the left, and the even powers will appear twice, to give:
2(2^5 + 10 \times 2^3u^2 + 5 \times 2u^4) = 464 64 + 160u^2 + 20u^4 = 464 u^4 + 8u^2 - 20 = 0We notice that -20 = -2 \times 10 and 8 = -2 + 10, so we can factor:
(u^2 - 2)(u^2 + 10) = 0and
u^2 = 2, \quad u^2 = -10So, we have four solutions:
u = \pm \sqrt{2}, \quad u = \pm i \sqrt{10}and so:
x=2 \pm \sqrt{2}, \quad y = 2 \mp \sqrt{2} x=2 \pm i\sqrt{10}, \quad y = 2 \mp i\sqrt{10}Example 2
a^2 - b = 241, \quad b^2-a = 241, \quad a \ne bSo:
a^2 - b = b^2 - a a^2 - b^2 - b + a = 0 (a + b)(a - b) + (a - b) = 0 (a + b + 1)(a - b) = 0So:
a + b = -1If a = -1/2 + u then b = -1/2 - u and:
(-1/2 + u)^2 - (-1/2 - u) = 241 (u^2 - u + 1/4) + 1/2 +u = 241 u^2 = 240 + 1/4 = 961/4 = 31^2/4So, u = \pm31/2 and a = 15, b = -16 (or a = -16, b = 15).
Alternatively:
a^2 + (a + 1) = 241 a^2 + a - 240 = 0We notice that -240 = -15 \times 16 and 1 = -15 + 16, so we can factor:
(a - 15)(a + 16) = 0with the same solutions as above.
Example 3
x + y = 10 xy = 34If x = 5 + u then y = 5 - u and:
(5 + u)(5 - u) = 34 25 - u^2 = 34u^2 = -9 = -3^2
So, we have two solutions u = \pm 3i and:
x = 5 \pm 3i, \quad y = 5 \mp 3iExample 4
A functional equation:
f\left(\frac {x} {x-1}\right) = 2f(x) + x^2, \quad x \ne 1This involves noticing that:
y = \frac {x} {x-1}, \quad x \ne 1 \iff x = \frac {y} {y-1}, \quad y \ne 1So:
f(x) = 2f\left(\frac {x} {x-1}\right) + \left( \frac {x} {x-1} \right)^2f\left(\frac {x} {x-1}\right) = \frac {1}{2}f(x) - \frac {1}{2}\left( \frac {x} {x-1} \right)^2
and:
0 = \frac {3}{2}f(x) + x^2 + \frac {1}{2}\left( \frac {x} {x-1} \right)^2or:
f(x) = -\frac {x^2} {3} \left( 2+ \frac {1}{(x-1)^2}\right) = -\frac {x^2} {3} \left( \frac {2x^2-4x+3}{(x-1)^2}\right)