How to determine what integer value of $j$ returns an integer when plugged in to $\sqrt{5j + 1}$ without trial and error? I am given $\sqrt{5j + 1}$ and I need to find what integer value of j evaluates this equation to an integer. I could only think of try and error by trying out different values of j incrementally to figure out the solution. Is there any other quicker way?

If we write $n^2=5j+1$, we can see that every value of $n^2 \equiv 1 \pmod 5$. That means $n^2$ must end in $1$ or $6$, which implies that $n$ ends in $1,4,6,$ or $9$.
But our choice of $n$ is free; for every $n \equiv 1,4,6,9 \pmod {10}$, some $j$ exists that satisfies the equation.
Hence, rearranging such that $j = \frac15 (n^2 -1)$, you can plug in any integer ending in one of those digits. For instance, for $n=1,4,6,9,11,14,16,19$, we find $j=0,3,7,16,24,39,51,72$.
For each residue class, we ought to be able to find an even simpler expression. If we look at $n \equiv n^2 \equiv 1 \mod{10}$, we see that $n^2 -1 = 100k^2 +20k = 5(20k^2+4k)$, and hence $j = 4k(5k+1), \forall k$, and of course $k$ is just the digits preceding $1$ in $n$.
For $n=1,11,21,31,41,51$, we have $k=0,1,2,3,4,5$, and $j=4k(5k+1)=0,24,88,192,336,520$.
You should be able to apply the same solution method for $n \equiv 4,6,9 \pmod{10}$.
EDIT: Adding some additional thoughts.
This could have been simplified by ignoring base $10$ completely, and working in modulo $5$ instead. Then we only have two cases, $n= \pm 1 \pmod 5$. This will be the case for solutions in squares with a given divisor, so if for instance we were examining $n^2=13k+1$, we would want to use modulo $13$.
This is great so long as we are working with squares and adding $1$ (or another square); once we no longer can factor $n^2-1=(n+1)(n-1)$, things will get a lot more difficult. If we were looking at $n^5 = 7k +3$, I'm not sure where the best place to start would even be.
