can the method of principle of mathematical induction be applied to prove or disprove all statements involving natural numbers?

Question

I was solving the following question

Let $S(k) = 1+3+5+\cdots+[2k-1] = 3+k^2$. Then which of the following is true ?

1. $S(1)$ is correct
2. $S(k) \implies S(k+1)$
3. $\neg (S(k) \implies S(k+1))$
4. Principle of mathematical induction can be used to prove the formula.

Clearly the first option is incorrect since $S(1) \implies 1 = 4$, which is false. Then I worked on second option . For this I assumed $S(k)$ is true, i.e., $1+3+5+...+[2k-1] = 3 + k^2$. Adding $(2k+1)$ on both sides we get, $$1+3+5+...+[2k-1]+[2k+1] = 3+k^2+2k+1 \implies 1+3+5+...+[2k-1]+[2(k+1)-1] = 3+(k+1)^2 \implies S(k+1)$$ is true , so option $2$ is correct and simultaneously option $3$ is incorrect . But when I tried to find the least value of $k$ for which $S(k)$ is true I couldn't find a single value of $k$. So how is this possible ?

Answer 1

$S(k)$ is false for all $k$. So $S(k)\implies$ absolutely anything. Hence $2$ is the only correct option.

I suspect that whoever set this question wants you to suppose that $S(k)$ is true and use elementary algebra to show that $S(k+1)$ is true, but that is logically invalid.

Also, you can't use induction to prove it, because it's false.

PS Your title seems to be asking a different question.