How can I show that the relation is not true for each natural number? Hello and a Happy New Year!!! 
Given the relation $w+nx$,where $w$ is a perfect square, $x>0$, $ n \in N$. I have to show that the relation $w+nx$ is not a perfect square for some $n$. So, since $w$ is a perfect square, it can be written as $w=b^2$. We consider that $w+nx$ is also a perfect square, so $w+nx=a^2⇒b^2+nx=a^2⇒n=\frac{a^2−b^2}{x}$. But how can I continue? How can I find something that is not true, so that we can say that $w+nx$ is not a perfect square for some $n$?
 A: The important part of your question is "how can I prove something is not a perfect square?".  There are actually several effective ways to do this, and it's extremely good practice to try some out and see which ones apply to your particular case (yes, there's usually more than one way to do it).  The key is to think about what's special about squares.  It turns out there's a lot of things that are special about squares, and almost any of them can be exploited to help you here (except one of them doesn't help and some are easier than others):


*

*Squares are never negative: therefore, if you could show that $w+nx$ is sometimes negative, then it can't always be a perfect square.

*Squares are pretty rare: there are only $k$ perfect squares between $1$ and $k^2$.  Therefore, if you could show that for some $k$, there are more than $k$ values of $w+nx$ between $1$ and $k^2$, then they can't all be perfect squares.

*Squares are well-spaced: given a perfect square $k^2$, the next smallest perfect square is $(k+1)^2$, so any square after $k^2$ is at least $2k+1$ away from $k^2$.  If you can find two values of $w+nx$ that are each at least $k^2$, but less than $2k+1$ apart, then they're too close to both be perfect squares.

*Squares only end in certain digits (in each base).  For instance, in base $10$, you might notice that squares always end in one of the digits $0,1,4,5,6$ or $9$, or that in base $7$ they always end in $0,1,2$ or $4$.  If you could show that some value of $w+nx$ ends with the digit $2$ (when written in base $10$) or ends in the digit $3$ (when written in base $7$), then that value cannot possibly be square.

*Squares only have even exponents in their prime factorization: for instance, the prime factorization of $20^2$ is  $(2^2\cdot 5)^2 = 2^{4} \cdot 5^2$.  If you could show that (for some value of $w+nx$) there's a prime $p$ that divides into $w+nx$ but where $p^2$ does not divide into $w+nx$, then that prime only occurs once in the factorization of $w+nx$, so it can't be square.

A: Assume, w + nx is a perfect square for any natural number n. Then, there is a number n, such that 
$$k^2 = w + nx$$ 
with x < 2k+1. Then 
$$k^2 < w + nx + x = k^2 + x < k^2 + 2k + 1=(k+1)^2$$.
Hence, w + (n+1)x cannot be a perfect square.
A: Here is an elementar number theory approach: 
Let $p>2$ be a prime not dividing $x$. Then $a^2$ can only have $\frac{p+1}{2}$ distinct residues $\bmod p$, because $a^2 \equiv (-a)^2 \bmod p$. As $b$ and $x$ are fixed and $x \not \equiv 0 \bmod p$, there are only $\frac{p+1}{2}$ destinct residues for  $(a^2-b^2)x^{-1} \bmod p$, too. Hence, not every natural number can be of this form.
