Bsrhrki

We know that the longest common substring of two strings can be found in O(N^2) time complexity.
Can a solution be found in only linear time?

Yes, the longest common substring of two strings can be found in $O(m+n)$ time, where $m$ and $n$ are the lengths of the two strings, assuming the size of the alphabet is constant.

Here is an excerpt from https://en.wikipedia.org/wiki/Longest_common_substring_problem.

The longest common substrings of a set of strings can be found by building a generalized suffix tree for the strings, and then finding the deepest internal nodes which have leaf nodes from all the strings in the subtree below it.

Building a generalized suffix tree for two given strings takes $Theta(m+n)$ time using the famous ingenious Ukkonen's algorithm. Finding the deepest internal nodes that come from both strings takes $Theta(m+n)$ time. Hence we can find the longest common substring in $Theta(m+n)$ time.

It is unlikely that that a better algorithm than quadratic exists, let alone linear. For the related problem of finding subsequences, this is a known result: In the paper "Tight hardness results for LCS and other sequence similarity measures." by Abboud et al. , they show that the existence of an algorithm with a running time of $O(n^2-varepsilon)$, for some $varepsilon>0$ refutes the Strong Exponential Time Hypothesis (SETH).

SETH is considered to be very likely true (although not universally accepted), so it is unlikely any $O(n^2-varepsilon)$ time algorithm exists.

While finding a substring is a slightly different problem, it seems likely to be equally hard.

Yes. There's even a Wikipedia article about it! https://en.wikipedia.org/wiki/Longest_common_substring_problem

In particular, as Wikipedia explains, there is a linear-time algorithm, using suffix trees (or suffix arrays).

Searching on "longest common substring" turns up that Wikipedia article as the first hit (for me). In the future, please research the problem before asking here. (See, e.g., https://meta.stackoverflow.com/q/261592/781723.)