A frequently used but false statement is
\begin{equation*}
(x+y)^n = x^n + y^n
\end{equation*}
This is sometimes referred to by mathematicians as the “child’s binomial theorem” (a quick trip to your search engine will turn up other names). One often sees examples of it such as
\begin{equation*}
\sqrt{x+y} = \sqrt{x}+\sqrt{y} \qquad \text{ and } \qquad
(x+y)^2 = x^2 + y^2.
\end{equation*}
While it is definitely false, there is something here that can be rescued. Notice that if we take \(x,y \in \mathbb{Z}\) and let \(n=2\text{,}\) then
\begin{equation*}
(x+y)^2 = x^2 + 2xy + y^2
\end{equation*}
and so if we look at everything modulo 2 we get
\begin{equation*}
(x+y)^2 \equiv (x^2 + 2xy + y^2) \equiv (x^2+y^2) \pmod 2.
\end{equation*}
Similarly, with \(n=3\) we have
\begin{equation*}
(x+y)^3 \equiv (x^3 + 3x^2y + 3xy^2 + y^3) \equiv (x^3+y^3) \pmod 3.
\end{equation*}
Indeed, one can show that for any prime number \(p\text{,}\) and integers \(x,y\) we have
\begin{equation*}
(x+y)^p \equiv x^p + y^p \pmod p.
\end{equation*}
Notice that this is not true for non-prime powers:
\begin{equation*}
(1+3)^4 = 4^4 = 256 \equiv 0 \pmod 4
\end{equation*}
while
\begin{equation*}
1^4 + 3^4 = 82 \equiv 2 \pmod 4.
\end{equation*}
