#ALgebra
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i mean 69 + 6 =/= 71
doing it by squaring both sides is also not that bad
ehh
yeah you get the same quadratic, -x^2 + 138x - 4761, the other way
so, conveniently, 138^2 = 4*4761
so the discriminant is 0
however good luck working that out
i dont like substituting y+x^2 type
my purpose was an aproach different than normal one
after all im giving the second toughest exam
anyways thank you
: D
dpp
i wonder if there is some very slick number theory way if you assume the answer is an integer
number theory scares me 💀
we need a^2 | x + 36, b^2 | (2x-33)
actually
x + 36 = na^2, 2x-33 = nb^2, 6x+6 = n(a+b)^2
n,b both odd, so a is also odd, so x is odd
lemme make this out
if x+36 = na^2
wouldnt 2x-33 = mb^2?
howd we know its the same factor for both (n)
the power of intuition
nah im not sure but i think if that is true it should be figureoutablewhy
we cant just assume it like that?
we cant just assume x is an integer either
did we?
.
my bad
shouldnt this work even when x is not an integer
| is only defined for integers, as is evenness/oddness
some stuff may still be workable if you only assume x is rational
ouh we studied this as if x|y x is divisible by y (x/y = integer)
doesnt matter if x and y are integers or not
thats a possibility
yeah its what im talking about
andy math ?
i aint watching his solution 💀
i was just working out for some unique trick workeable on any question im stuck on
just boosting my solving capabilities
n,x are both 1 mod 4
3x = -1 mod 16 (from last equation, n == 1 mod 4, (a+b)^2 == 4 mod 4)
x = 5 mod 16
na^2 = nb^2 = 9 mod 16
so a = b or a = -b mod 16
is mod the same thing as it is in programming
remainder thing
yea thats it
yeah pretty much
however if 16 | (a+b) then the lhs is a multiple of 256
meanwhile, 6x + 6 = 6(4k + 2) = 12(2x + 1)
which is not a multiple of 256
so a == b mod 16
anyway, at this point i think this is harder than the normal method(s), but you can try doing it this way kind of like a puzzle
+close