Is my hypothesis correct? $$\left| \left|(a^2) - 25\right|-b\right| + b = 0$$
You have to prove that $b<0$ and $b=0$ at the same time 
I have no problem to prove that $b$ can be $0$ the thing that I need help with is $b<0$. My thought there is a bit complicated and some people I've asked have said that's it's not valid but here it is. Oh and by the way sorry that I don't write pure mathematics I am a rookie just a 16 year old kid  so my thoughts will more on like the in the form of words(I know that it's not 100% maths but these are my thoughts). Sorry if it sounds silly.So here it starts.
Let's break the whole thing down into 2 parts
1)$\left| \left|(a^2) - 25\right|-b\right|$
2)$+ b$
As it's well know in the subtraction of two numbers in order to be $0$ (considering that they belong to Real)numbers they should be opposite for ex.
$4-4=0$ or $a-a=0$
That means that one of them should be negative and another positive AND THEY SHOULD BE REPRESENTED BY THE SAME SYMBOL BUT OTHER SIGN
Let's hypothesize that $b$ is actually smaller that $0(b<0)$. Each number consists from it's number sign(+,-) and an number sign and a symbol(1,2,3,4,5....)(Again I am talking about Real numbers). Let's call the symbol "$y$" because of it's randomness so b should be equal with (- "$y$")
The thing now changes from :
$\left|\left|(a^2) - 25\right|-b\right| + b$  to :
$\left|\left|(a^2) - 25\right|-(- y)\right| + (-y) $
since $b$ is negative it should be represented as ($- y$) because of it's negativity
So it proves out that :
$\left|\left|(a^2) - 25\right| + y\right| - y$
That guides us to think that the negative part on this is the second part actually because the second part ( $- y$) is negative the first one should be the positive one a fact that can be verified from the fact that $\left|\left|(a^2) - 25\right| + y\right|$ is a positive number since $|-4|=|4|=4$ or $|y|=|-y|=y$
Because the 1st part and the 2nd part are represented by the same symbol but not sign we supposed that
$\left|\left|(a^2) - 25\right| + b\right| = + y$ like the positive part 
and 
$+ b = - y$ the negative part
this means that
$b + b \Rightarrow + y - y = 0 \Rightarrow y=y$ something that is valid so 
I hypothesized something and it turned out that my result is something valid this must mean that my hypothesis is correct
 A: $||a^2 - 25| - b| + b = 0 \implies$
$||a^2 - 25| - b| = -b$.
But $||a^2 - 25| - b| \ge 0$ because absolute values are non negative.
So $-b \ge 0$ so $b \le 0$  (i.e.  $b$ is negative or 0.)
And that's it.  We are done.  We know if the equation is true $b \le 0$.
But we don't know if there is more restrictions we can have and we don't know if the equation itself is even possible.
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I want to go further and see if there is anything more we can deduce.
Now $|a^2 - 25| \ge 0$ and $b \le 0$ so $|a^2 - 25| - b \ge 0$. 
So $||a^2 - 25| - b| = |a^2 - 25| - b$.
So $-b = ||a^2 - 25| -b| = |a^2 - 25| -b$
So $|a^2 - 25| = 0$
So $a^2 = 25$ and $a = \pm 5$.
$b \le 0$ and can be any such value.
You can not prove it is BOTH $b < 0$ and $b = 0$.  That would be a contradiction but $b < 0$ OR $b = 0$.  This are individually both possible. and any possible value that is either less than or equal to 0 is acceptable.
A: If $b>0$, the number
$$
\bigl|\,|a^2-25|-b\bigr|+b
$$
is strictly positive. So if you want it to be $0$, you need $b\le0$.
Suppose $b\le0$ and $\bigl|\,|a^2-25|-b\bigr|=-b$. Then squaring gives
$$
|a^2-25|^2-2b|a^2-25|+b^2=b^2
$$
so either $|a^2-25|=0$ or $|a^2-25|-2b=0$. In the latter case, $2b=|a^2-25|\ge0$, which forces $b=0$ and $|a^2-25|=0$.
Thus the equality holds for $a=\pm5$ and $b\le0$.
