MHB Solve Logic Puzzle: Predicates & True/False Explained

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The discussion centers on understanding predicates and true/false statements in logic. The first example is identified as a statement with its negation being true, indicating the original statement is false. The second example is recognized as a predicate due to the unbound variable, and a suggestion is made to express it in a universally quantified form for negation. The third example is confirmed as a true statement, with a reference to a theorem by Fermat regarding primes and sums of squares. Clarification on predicates involves noting that they contain unbound variables, distinguishing them from definitive true/false statements.
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I'm unsure about these three, here are my attempts. Please also explain the difference between a predicate and true/false. I assumed it is a predicate when it can be either true or false.

a) Predicate. Negation is ¬(∃n ∈ N n²>n)
b) True. Negation is, "When x<0 there is y such that y^2=x
c) No clue :P

Your help is truly appreciated!
 

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I assume by predicate, it is meant that the sentence has unbound variables. So

a) This is a statement. The negation is
$$\exists n \text{ such that }n\in N \text{ and }n^2\leq n$$
The negation is true (set n=1) and so the original statement is false.

b) The variable x is not bound, so this is a predicate. To make a statement one might say:
$$\forall x \in R\text{ if }x<0 \text{ then }\not\exists y\in R \text{ such that }y^2=x$$
I leave it to you to negate this statement.

c) This is a true statement. ("famous" theorem of Fermat says a prime is the sum of two squares iff the prime is congruent to 1 mod 4)
The negation is:
$$\exists p\in P \text{ such that }\exists n\in N \text{ with }p=4n+1\text{ and }\forall a\in N\,\forall b\in N\,a^2+b^2\neq p$$
 
I was reading a Bachelor thesis on Peano Arithmetic (PA). PA has the following axioms (not including the induction schema): $$\begin{align} & (A1) ~~~~ \forall x \neg (x + 1 = 0) \nonumber \\ & (A2) ~~~~ \forall xy (x + 1 =y + 1 \to x = y) \nonumber \\ & (A3) ~~~~ \forall x (x + 0 = x) \nonumber \\ & (A4) ~~~~ \forall xy (x + (y +1) = (x + y ) + 1) \nonumber \\ & (A5) ~~~~ \forall x (x \cdot 0 = 0) \nonumber \\ & (A6) ~~~~ \forall xy (x \cdot (y + 1) = (x \cdot y) + x) \nonumber...
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