Find an equation where all 'y' is always irrational for all integer values of x Intuitively it appears to me that if $x$ is an integer, $y$ has to be an irrational number for the following equation.
$10y^2-10x-1 = 0$
Can someone prove me right or wrong? 
 A: You are correct. 
If $$10y^2-10x-1 = 0$$ and if $x$ is an integer then $y$ is irrational. 
Note that if $y= \frac {m}{n}$  where $m$ and $n$ are relatively prime, then $$ 10 m^2=n^2(10x+1)$$
Obviously $10x+1$  is not a multiple of $10$ which makes $n^2$ to be a multiple o f $10$. Now if $n^2$ is a multiple of $10$ , $n$ is a multiple of $10$ thus $n^2$ is a multiple of $100$ which in turn makes $m $ to be a multiple of $10$
That contradicts the relatively prime assumption of $m$ and $n$.
Thus, $y$ is irrational.    
A: If $y$ is a rational number then $y$ should be an integer number, but $10x+1$ is not divisible by $10.$
We got a contradiction.
A: You are correct.  To see this, suppose you had a counterexample for $x=n\in \mathbb Z$ and write $y=\frac ab$ with $a,b\in \mathbb N$ with $\gcd(a,b)=1$.  Then we see that $$10a^2=b^2(10n+1)\implies 10\,|\,b^2$$ which contradicts the relative primality of $a,b$.
Perhaps a simpler example would be $y^2=4n+3$ . It is a general result that this can not have a rational solution unless it has an integral solution, and there is no integer which when squared is $3$ more than a multiple of $4$.
A: Suppose that the equation has a quotient answer $y=\frac{p}{q}$ where $g.c.d.(p,q)=1$ therefore by substitution we get:$$\frac{p^2}{q^2}=\frac{10x+1}{10}$$or equivalently$$10p^2=(10x+1)q^2$$ since $p^2$ and $q^2$ are coprime and so are $10$ and $10x+1$ we can conclude that:$$10|q^2\\q^2|10$$
this leads to $q^2=10$ or $q=\sqrt 10 \in \Bbb Z$ which is impossible. Then we have proven that all of roots of the equation are irrational for $x\in \Bbb N$
