Expressions that equate to a square $4x+1 = a^2$
Where a and x is an integer, what are all possible value of x?
I always encounter these types of question even in the form of 
$ax^2 + bx + c$
How do I approach these kinds of questions?
 A: This is quite obvious that $a$ is odd, therefore there exists some integer $k$ such that $$a=2k+1$$and$$a^2=4k^2+4k+1=4(k^2+k)+1=4\underbrace{k(k+1)}_{x}+1$$therefore for $x$ being as product of two consecutive integers $a$ is also an integer so $a$ can be any odd integer.
A: You are looking for $a$ such that the remainder of dividing $a^2$ by $4$ is $1$ (then $x$ is the quotient). The possible remainders when dividing by $4$ are $0$, $1$, $2$, and $3$. Suppose the remainder of dividing $a$ by $4$ is $r$. Then the remainder for $a^2$ is directly related to $r^2$: $a=4x+r$ implies $$a^2=(4x+r)^2=16x^2+8x+r^2=4(4x^2+2x)+r^2,$$
where if $r^2>4$ then we should extract factors of 4 from it (and add them to the quotient) until it turns below 4, in order to get a true remainder.
If $r=0$ then $r^2=0$; if $r=1$ then $r^2=1$; if $r=2$ then $r^2=4$, which changes to $0$ (by adding 1 to the quotient) since $r^2<4$. If $r=3$ then similarly $r^2=9$ changes to $1$. Therefore the remainder of $a^2$ is $1$ if and only if the remainder of $a$ is either 1 or 3, $a=4c+1$ or $a=4c+3$ for some $c$: 1 and 3 (c=0), 5 and 7 (c=1), etc.  So the solutions are all odd integers, since $a=4c+1$ or $a=4c+3$ is equivalent to $a=2d+1$ for some $d$.
By the way, if you are interested in this kind of reasoning, you sould learn about modular arithmetic, which studies the algebra of remainders. In this example we would say that $a^2\equiv 1(\mod 4)$ implies $a\equiv 1,3(\mod 4)$ and also that $r\equiv 9\equiv 1(\mod 4)$.
A: If $x$ is also an integer, then:
$4x+1 = a^2 \to x=\dfrac{a^2-1}{4}\to x=\dfrac{(a-1)(a+1)}{4}$
Now if $a$ is odd, then $a-1$ and $a+1$ are even thus their product is divisible by $4$ and $x$ will be an integer too.
