If $n^2+4n+10$ is a perfect square, then find the possible integer values of $n$. 
If $n^2+4n+10$ is a perfect square, then find the possible integer values of $n$.

I couldn't understand what the question is asking me to do. I could only do one step that would equate it to $k^2$ after that I wasn't able to solve it.
 A: Hint. What is $ ( n + 2 ) ^ 2 $?
A: Let $$n^2+4n+10 = k^2$$ 
Then 
$$k^2-(n+2)^2=6$$
$$\implies k^2=6+(n+2)^2$$
Since $$k^2 \equiv  0~ \text{or} ~1 \pmod4$$
And ,$$6+(n+2)^2  \equiv  2~ \text{or} ~3 \pmod4$$
No solution!
A: Write $n^2+4n+10 = m^2$. Then $6=m^2-(n+2)^2=(m+n+2)(m-n-2)$.
Now $6=ab=(\pm 1)(\pm 6)$ and $6=(\pm 2)(\pm 3)$ are the only possible factorizations of $6$.
Therefore, we need to solve $m+n+2 = a, m-n-2=b$ for $m,n$.
However, the determinant is $-2$ and we'd need $a,b$ to have the same parity, which never happens. Therefore, there are no integer solutions.
The same argument proves that an integer is the difference of two integer squares iff it is odd or a multiple of $4$.
A: Clearly there are no positive integer solutions because we see from the inequality below that $k^{2}= (n+2)^{2}+6$ lies strictly between two consecutive squares:
$(n+2)^{2} < k^{2}<(n+3)^{2}$
It's easy to manually check that $n=-1,-2,-3,-4$ also fail to make $(n+2)^{2}+6$ a perfect square. The absence of all other negative integer solutions is similarly seen by setting $n=-m,$ for $ m \in \mathbb{N}$ and by noting that $\forall \, m\geq 5$ we have:
$(m-2)^{2}<(m-2)^{2}+6<(m-1)^{2}$
A: If $ax^2 + bx + c$ is a perfect square, then there exists only one solution to
$ax^2 + bx + c = 0$ as it can then be written as $(x + p)^2$ for some p. This means $b^2 = 4ac$. But in the given equation it is not true ($4^2 != 4*10$), therefore the given equation can never be a perfect square for any real n.
A: For integers $n,k$  we have $$n^2+4n+10=k^2\implies |k|^2=6+|n+2|^2\implies$$  $$ \implies |k|>|n+2|\implies |k|\geq 1+|n+2|\implies$$  $$\implies 6=|k|^2-|n+2|^2\geq (1+|n+2|)^2-|n+2|^2=1+2|n+2|\implies$$  $$ \implies 6\geq 1+2|n+2|\implies 5\geq 2|n+2|\implies$$  $$\implies 2+\frac {1}{2}\geq |n+2|\implies 2\geq |n+2|\implies$$ $$ \implies |n+2| \in \{0,1,2\} \implies  |n+2|^2\in \{0,1,4\}\implies$$ $$\implies|k|^2=6+|n+2|^2\in \{6,7,10\}\implies 1=0.$$
