How do Quadratic Residues work and how do they relate to Diffie Hellman? Can someone explain Quadratic residues to me in english? I keep reading forums with all the math symbols and it's hard to follow. And how are they incorporated in Diffie-Hellman? Is there a difference in using them vs cubic residues?
Can someone explain Quadratic residues to me in english?


A value $x$ is a Quadratic residue modulo $p$ if there exists a value $t$ for which $t^2 \bmod p = x$.

That is, we take $t$, square it (that's the Quadratic part), and then take it modulo $p$ (that's the residue part, that is, the part that's left over after removing all the multiples of $p$), and that gives us the value $x$.  If such a process can give us $x$, we say that $x$ is a Quadratic Residue.

For example, $2$ is Quadratic Residue modulo $7$ (because if we start with $3$, square it, and then take that result modulo $7$, we end up with $2$); however $3$ is not (can be verified by computing $1^2 \bmod 7, 2^2 \bmod 7, 3^2 \bmod 7, …, 6^2 \bmod 7$)

Also, in this answer, I have been careful to always say "Quadratic Residue modulo $p$"; obviously, the same value $x$ can be a Quadratic Residue modulo some prime $p$ and not modulo another prime $q$.  Normally, you'll see the statement "$x" is a Quadratic Residue"; they mean modulo something, but that something is implicit (as they are working in a specific group, which should be obvious from context).

Important facts:


If $p$ is a prime greater than 2, then some of the values between $1$ and $p-1$ will be Quadratic Residues modulo $p$, and some will not be.  In fact, precisely half of those values will be and half will not.
If $p$ is a prime, then $a \times b \bmod p$ will be a Quadratic Residue modulo $p$ if either both $a$ and $b$ are Quadratic Residues, or both are not.
If $p$ is a prime, it turns out to be simple to check where $x$ is a Quadratic Residue mod $p$.



  And how are they incorporated in Diffie-Hellman?


Mostly, it's not.  If the Diffie-Hellman generator $g$ is not a Quadratic Residue modulo $p$, then given $g^a \bmod p$, we can determine whether $a$ is even or odd; given $g^a \bmod p, g^b \bmod p$, we can determine whether $g^{ab} \bmod p$ is a Quadratic Residue modulo $p$ or not.  On the other hand, we generally perform Diffie-Hellman using a generate that generates a large prime subgroup; such a generator will always be a Quadratic Residue modulo $p$, and so examining the values $g^a \bmod p, g^b \bmod p$ doesn't tell us anything.


  Is there a difference in using them vs cubic residues?


If $p$ is a prime where $p \equiv 2 \bmod 3$, then cubic residues modulo $p$ are uninteresting; all values $x$ are Cubic Residues modulo $p$.

If $p$ is a prime with $p \equiv 1 \bmod 3$, then a third of the values between $1$ and $p-1$ will be cubic residues modulo $p$, and two thirds will notbe.  On the other hand, the same Diffie-Hellman logic applies; if $g$ generates a large prime subgroup, then $g$ will be a cubic residue modulo $p$, and so will $g^a \bmod p, g^b \bmod p, g^{ab} \bmod p$ as well...