# Finding integer solution in a set

I want a number theoretic answer to the below question, taken from the section 1.3, question no. #22 from the book of 'Abstract Algebra' by 'Hillman, Peterson'.

Let $c$ be in $1+\mathbb{Z}^{+}=\{2,3,4,\ldots\}$. Find one solution in positive integers $x$ and $y$ to the equation $x^2-y^2=4c$.

I have tried the below approach without success.

Let $x-y=1$, then get equation: $x+y=4c$.

Now $x=(4c+1)/2$, $y=(4c-1)/2$. But, obviously these are not integer values as a factor of $\pm1/2$ is there.

I am not sure of, but can I take any value for $x-y$ above, say $2$; so as to make the final value of $x$ and $y$ an integer.

That way, $x-y=2$, then $x=2c+1$, $y=2c-1$.

I hope that this is a trivial question and any value in the given set of values ($1+\mathbb{Z}^{+}$) can be taken to be $x-y$; and the choice depends on the problem and the desired solution.

However, on changing the question a little — if a given value is given instead of $4c$, then how it can be proved that this value cannot (if not) be found from the set; e.g. the value $34$.

So, how it can be proved that $x^2-y^2=34$ has no integer value solution in the given set, or the general case of the set $\mathbb{Z}$.

## 1 Answer

The OP asks how it can be proved that $x^2-y^2=34$ has no integer solutions. One way is to look mod $8$.

All squares mod $8$ belong to $\{0,1,4\}$. So the difference of two squares mod $8$ belongs to $\{0,1,3,4,5,7\}$. Since $34\equiv2$ mod $8$, it cannot be the difference of two squares.

• But, will this work for mod 5, where the set of values for squares mod 5 is {0,1,4}. The difference of squares will have residue in the set {0,1,3,4}. Also, 34 mod 5 is 4. So, is the logic of residue class working still, or I am wrong in reasoning somewhere. – jiten Nov 2 '17 at 2:31