Advanced Math Help with Any Problems!

Suppose that a and b are real numbers with $0<a<1$ and $0<b<1$. Prove by contradiction or contrapositive: If $a^2+b^2=1$, then $a+b>1$.
So far I get the below:
To prove the above statement with contrapositive, we need to show that if $a+b\le <!--\; \le \; -->1$ then $a^2+b^2\ne <!--\; \ne \; -->1$. If $a+b\le <!--\; \le \; -->1$, then $a\le <!--\; \le \; -->1-<!--\; -\; -->b$. If we square both sides of the inequality we get $a^2\le <!--\; \le \; -->(1-<!--\; -\; -->b{)}^2$, which is $a^2\le <!--\; \le \; -->1-<!--\; -\; -->2b+b^2$, then $a^2-<!--\; -\; -->b^2\le <!--\; \le \; -->1-<!--\; -\; -->2b$.
I get to this step but I get $a^2-<!--\; -\; -->b^2$ instead of $a^2+b^2$, I don't know if I am on the right track or if it is possible to prove with contrapositive?

Condition making a relation reflexive
The relation $R=(a,b)\mid a=b\text{or}a=-b[a,b\in Z]$ is reflexive, as per Discrete Mathematics and Its Applications 8th Edition by Kenneth Rosen.
Is this because $a=b$ short-circuits the evaluation of $a=b$ or $a=-b$ (making it always true), or is there some other reason why the relation is reflexive?

Linear Diophantine equation with three variables and a condition
Let me start by saying that I'm new to Diophantine equations and my method certainly will not be the best possible one. I'm aware that a faster method of solving this particular case exists, but I want to know if what I did was correct and how I can finish the exercise.
$39x+55y+70z=3274$
$x+y+z=69$
$x\ge <!--\; \ge \; -->0,y\ge <!--\; \ge \; -->0,z\ge <!--\; \ge \; -->0$
What I tried:
$39x+55y=3274-<!--\; -\; -->70z$
The GCD for 39 and 55 is one, so we can set aside $z=t$ to be a parameter, where t is an integer. We proceed to solve this as a Diophantine equation with two variables.
I now need to use the extended Euclidean algorithm to express 1 as a linear combination of 39 and 55.
I got that $1=24\cdot <!--\; \cdot \; -->39-<!--\; -\; -->17\cdot <!--\; \cdot \; -->55$
Now, the solution would be
$x=78576-<!--\; -\; -->1680t+55s$
$y=-<!--\; -\; -->55658+1190t-<!--\; -\; -->39s$
$z=t$
where t and s are integers. I got this from the formula that $x={\mathrm{\lambda <!--\; \lambda \; -->}}_1\cdot <!--\; \cdot \; -->b+s\cdot <!--\; \cdot \; -->a_2$ and $y={\mathrm{\lambda <!--\; \lambda \; -->}}_2\cdot <!--\; \cdot \; -->b-<!--\; -\; -->s\cdot <!--\; \cdot \; -->a_1$.
Where ${\mathrm{\lambda <!--\; \lambda \; -->}}_1$ and ${\mathrm{\lambda <!--\; \lambda \; -->}}_2$ are the coefficients in the Euclidean algorithm, b is $3274-<!--\; -\; -->70t$ and $a_1$ and $a_2$ are 39 and 55, respectively.
It's obvious that $t\ge <!--\; \ge \; -->0$, and if I put that x and y are greater than zero, I get that
$s\ge <!--\; \ge \; -->\frac{1680t-<!--\; -\; -->78576}{55}$ and $s\le <!--\; \le \; -->\frac{-<!--\; -\; -->55658+1190t}{39}$ which makes $\frac{1680t-<!--\; -\; -->78576}{55}\le <!--\; \le \; -->\frac{-<!--\; -\; -->55658+1190t}{39}$ which is $t\le <!--\; \le \; -->46.77$.
I now have the whole range of integers [0,46] which of course isn't feasible to do (as I'd have to plug them into the inequalities for s and get even more cases.
How do I proceed from here? What do I do? I still have the condition $x+y+z=69$ but I feel like I'm missing something. Can it be really done this way, except that it's an unimaginably hefty job of testing each case?

How many different triples $f:A\to <!--\; \to \; -->S,g:B\to <!--\; \to \; -->S,h:C\to <!--\; \to \; -->S$ are there such that f, g, h are all injective, and $f(A)=g(B)=h(C)$?
In this Question we know that the $|A|,|B|,|C|=m>0$ and set S has cardinality $n>0.$.
Since the functions are one-to-one this means each domain maps a codomain and as all three sets (Domain) have same cardinality we can deduce that ${}_n{C}_m$ possible codomain mapped by these functions.
And because these functions are One-One, so each function can form ${}_n{P}_m$ number of One-One function so according to me this leads me to ${}_n{P}_m/{}_n{C}_m$ as number of different triples that can be formed.

Prove by Mathematical Induction ${𝑎}_{𝑛}<3𝑥,$, if ${𝑎}_1=3,𝑥\ge 2,$, and ${𝑎}_{𝑛+1}=\sqrt{2(2+{𝑎}_{𝑛})}$

Stuck on this question
Water fills a tank at an increasing rate of 50 litres during the first hour, 75 litres during the second hour, 100 litres during the third hour and so on. The size of the tank is $16\times <!--\; \times \; -->6\times <!--\; \times \; -->3$. All dimensions in metres. i am stuck trying to figure out how long it takes to fill the tank to the nearest hour.

Deducing Inequality from an Equation
I was interested in Solution of this Non-Homogenous Recurrence Relation
$f(n)=f(n-<!--\; -\; -->1)+f(n-<!--\; -\; -->3)+1$
The Base conditions are:
$f(0)=1$
$f(1)=2$
$f(2)=3$
Since, the equation is non-homogenous, it will consists of Two Parts.
- Finding solution of Associated Homogenous Recurrence Relation $f(n)=f(n-<!--\; -\; -->1)+f(n-<!--\; -\; -->3)$
For this the characteristic equation will be $x^3-<!--\; -\; -->x^2-<!--\; -\; -->1=0$
- Finding Particular Solution
one can find answer by inputting $f(n)=f(n-<!--\; -\; -->1)+f(n-<!--\; -\; -->3)+1,f(0)=1,f(1)=2,f(2)=3$
The solution is quite complex. Thus, I was interested if we can deduce that $f(n)\ge <!--\; \ge \; -->x^n$ for some $x\mathrm{ϵ<!--\; ϵ\; -->}\mathbb{R}$.

Find k: ${27}^k\equiv <!--\; \equiv \; -->2\mathrm{mod}2021$
By using these: $2^{11}=2048\equiv <!--\; \equiv \; -->27\mathrm{mod}2021$
$2021=43\cdot <!--\; \cdot \; -->47$
I should find integer k
${27}^k\equiv <!--\; \equiv \; -->2\mathrm{mod}2021$
Well, I can only come up with something like this:
$a^{42}\equiv <!--\; \equiv \; -->1\mathrm{mod}43$
$b^{46}\equiv <!--\; \equiv \; -->1\mathrm{mod}47$
$a\ne <!--\; \ne \; -->0\mathrm{mod}43,b\ne <!--\; \ne \; -->0\mathrm{mod}47$

Need a clarification of the proof that the prime ideal space of a distributive bounded lattice is compact
Let L be a bounded distributive lattice, then the prime ideal space $⟨<!--\; ⟨\; -->{\mathcal{I}}_p(L);\mathrm{\tau <!--\; \tau \; -->}⟩<!--\; ⟩\; -->$ is compact. $\mathrm{\tau <!--\; \tau \; -->}$ is the topology whose basis is $\mathcal{B}=\{X_b\cap <!--\; \cap \; -->(X\setminus <!--\; \setminus \; -->X_c):b,c\in <!--\; \in \; -->L\}$, with $X_a=\{I\in <!--\; \in \; -->{\mathcal{I}}_{\mathcal{p}}(L):a\notin <!--\; \notin \; -->I\}$.
We want to prove that the subbasis $\mathcal{S}=\{X_b:b\in <!--\; \in \; -->L\}\cup <!--\; \cup \; -->\{X\setminus <!--\; \setminus \; -->X_c:c\in <!--\; \in \; -->L\}$ satisfies Alexander's Lemma. Let $\mathcal{U}:=\{X_b:b\in <!--\; \in \; -->A_0\}\cup <!--\; \cup \; -->\{X\setminus <!--\; \setminus \; -->X_c:c\in <!--\; \in \; -->A_1\}$, a open cover of ${\mathcal{I}}_{\mathcal{p}}(L)$.
Let J be the ideal generated by $A_0$ and G the filter generated by $A_1$. It is easy to prove that $J\cap <!--\; \cap \; -->G\ne <!--\; \ne \; -->\mathrm{\varnothing <!--\; \varnothing \; -->}$. So $J\cap <!--\; \cap \; -->G\ne <!--\; \ne \; -->\mathrm{\varnothing <!--\; \varnothing \; -->}$ and let $a\in <!--\; \in \; -->J\cap <!--\; \cap \; -->G$; if $A_0$ and $A_1$ are both non-empty, there exist $b_1,\dots <!--\; \dots \; -->,b_j\in <!--\; \in \; -->A_0$ and $c_1,\dots <!--\; \dots \; -->,c_k\in <!--\; \in \; -->A_1$ s.t. $c_1\wedge <!--\; \wedge \; -->\cdots <!--\; \cdots \; -->\wedge <!--\; \wedge \; -->c_k\le <!--\; \le \; -->a\le <!--\; \le \; -->b_1\vee <!--\; \vee \; -->\cdots <!--\; \cdots \; -->\vee <!--\; \vee \; -->b_j$, whence $X=X_1=X_{b_1}\cup <!--\; \cup \; -->\cdots <!--\; \cdots \; -->\cup <!--\; \cup \; -->X_{b_j}\cup <!--\; \cup \; -->(X\setminus <!--\; \setminus \; -->X_{c_1})\cup <!--\; \cup \; -->\cdots <!--\; \cdots \; -->\cup <!--\; \cup \; -->(X\setminus <!--\; \setminus \; -->X_{c_k})$
What escapes me is why we can write 1 in that way if $J\cap <!--\; \cap \; -->G\ne <!--\; \ne \; -->\mathrm{\varnothing <!--\; \varnothing \; -->}$?

How to calculate summation of $\frac{n^2+n+1}{n^2+n}$ over 1 to 25?
Can we calculate the summation of numerator, and denominator and then divide them both? But that does not seems to be a plausible way. So, I reduced it to $S=1+\frac{1}{n^2+n}$. But now I do not how to compute this sum.

Find smallest positive $a\in <!--\; \in \; -->\mathbb{Z}$ such $10a\equiv <!--\; \equiv \; -->7\mathrm{mod}17$
I found that the answer was $a=16$, I believe.
Note that $10a\equiv <!--\; \equiv \; -->7\mathrm{mod}17⟺<!--\; ⟺\; -->17\mid <!--\; \mid \; -->10a-<!--\; -\; -->7⟺<!--\; ⟺\; -->10a-<!--\; -\; -->7=17x,x\in <!--\; \in \; -->\mathbb{Z}$
The way I went about it was running EEA for $10a+17(-<!--\; -\; -->x)=7,x\in <!--\; \in \; -->\mathbb{Z}.$.
I found it to be $1=10(-<!--\; -\; -->5)+17(3)⟹<!--\; ⟹\; -->7=10(-<!--\; -\; -->35)+17(21)$. But I wasn't sure where then I could find 16 for this. I ended using bruteforce to find $a=16$ starting from 1. (not efficient).
Any more efficient alternative to find this such a is appreciated!

Binary strings with exactly n ones but without 000 or 111
I am a newbie in binary strings and generating series. I have a problem of enumeration which can be transformed to the following: find the number of binary strings of exactly n ones which do not contain three consecutive ones or three consecutive zeros. The answer can be given as a coefficient of a generating series.

How to solve $x(x-<!--\; -\; -->1)(x-<!--\; -\; -->2)(x-<!--\; -\; -->3)(x-<!--\; -\; -->4)(x-<!--\; -\; -->5)....(x-<!--\; -\; -->999)$

How do I prove $B=C$ when...
If $A\cap <!--\; \cap \; -->B=A\cap <!--\; \cap \; -->C$ and $A^{\mathrm{\complement <!--\; \complement \; -->}}\cap <!--\; \cap \; -->C$, then $B=C$.
I seem to be able to prove this without the use of $A^{\mathrm{\complement <!--\; \complement \; -->}}\cap <!--\; \cap \; -->C$ but it must be there for a reason. Where would I add this in. The basic gist behind what I did was say x was an element of A therefore x is in $A\cap <!--\; \cap \; -->B$ and $A\cap <!--\; \cap \; -->C$. So x is in B and x is in C. So since they share the same element they are equal. This may not be right but I think I'm somewhere in the ballpark. Any help would be greatly appreciated.