SMOJ wrote: Let $n$ be a positive integer, and let $x=\frac{\sqrt{n+2}-\sqrt{n}}{\sqrt{n+2}+\sqrt{n}}$ and $y=\frac{\sqrt{n+2}+\sqrt{n}}{\sqrt{n+2}-\sqrt{n}}$. It is given that $14x^2+26xy+14y^2=2014$. Find the value of $n$. AnswerANSWER:$\boxed{n=5}$ SolutionFirstly, we divide $14x^2+26xy+14y^2=2014$ it by $2$. Then $7x^2+13xy+7y^2=1007$. Now observe that $x=\frac{1}{y}$, so $x\times y=1$. Hence $7x^2+13xy+7y^2=1007$ becomes $7x^2+13+7y^2=1007$ which implies that $7(x^2+y^2)=1007-13=994$. So, we have $(x^2+y^2)=142$ or that $(x^2+{\frac{1}{{x}^{2}}})=142$ or that $(x^2+{\frac{1}{{x}^{2}}}+2)=144={12}^{2}$. So we have that $x+\frac{1}{x}=\pm 12$. Now observe that since $x=\frac{\sqrt{n+2}-\sqrt{n}}{\sqrt{n+2}+\sqrt{n}}$ and $y=\frac{\sqrt{n+2}+\sqrt{n}}{\sqrt{n+2}-\sqrt{n}}$ we can rationalize them as $x= \frac{n+2+n-2\sqrt{n(n+2)}}{2}=n+1-\sqrt{n(n+1)}$ and similarly $y=n+1+\sqrt{n(n+1})=\frac{1}{x}$ hence we now have that: $x+\frac{1}{x}=n+1-\sqrt{n(n+1)}+n+1+\sqrt{n(n+1)}=2(n+1)=\pm12$. Since $n$ is a positive integer, we have $n+1=\frac{12}{2}$ or that $n=5$.

We first multiply the given equation by $\dfrac{7}{2}$ to obtain $49x^2 + 91xy + 49y^2 = 7049$ $\implies (7x + 7y)^2 - 7xy = 7049$. It is easy to see that $xy = 1$. Hence we have $(7x + 7y)^2 = 7056$ $\implies 7x + 7y = \pm 84 \implies x + y = \pm 12$. By cross-multiplying, we see that $x + y$ $= \dfrac{(n + 2) + n - 2\sqrt{n}\sqrt{n + 2} + (n + 2) + n + 2\sqrt{n}\sqrt{n + 2}}{(n + 2) - n} = \dfrac{4n + 4}{2} = 2n + 2$. Hence $n + 2 = \pm 12 \implies n = 5, -7$. However, $n = -7$ would lead to square roots not defined across the reals in our initial equation. Therefore $n = 5$.

Identify that $y=\frac{1}{x}$ and it's converse. Using this piece of information, rewrite that terrible equation to become $$x^2+\frac{1}{x^2}=142$$Then find the conjugates of $x$ and $y$ and multiply them together. Therefore we have $x^2$ and $y^2$ equal to some very long polynomials. However, going back to that first piece of information we can simplify this equation tremendously to the point that $x^2+y^2=142$. After you substitute those long polynomials for $x^2$ and $y^2$, you should get something like this. $$142=2(n+1)^2+2n(n+2)$$which simplifies to this nice and tasty polynomial $$4n^2+8n+2=142$$. Now we can simplify this equation into this $$4n^2+8n-140$$and divide the entire polynomial by $4$ into $$n^2+2n-35$$. $n^2+2n-35$ factors into $$(n+7)(n-5)$$which has two roots; $n=-7$ and $n=5$. However, since $n$ can't be negative, we know that our only solution is $n=5$.

$(x^2+\frac{1}{x^2})=142$