Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

enter image description hereThe question is to find the number of solutions such that $(x, y)$ are integers: $(x-8)(x-10)=2^y$. Here's what I did: $u(u-2)=2^y$. From the quadratic formula, $u=1+\sqrt{1+2^y}$. This is where I think this question was meant to be solved by modular arithmetic here, but I don't know it so please try not to use it in your answers. I now constructed two right angle triangles. Triangle 1 has sides $1$, $2^{y/2}$,a hypotenuse of $\sqrt{1+2^y}$, and an angle $t$ such that $\sin(t)=\dfrac {1}{\sqrt{1+2^y}}$. Triangle 2 has sides of $1$, $w$, a hypotenuse of $u$, and an angle $t$ such that $\cos t=\dfrac{1}{u}$. Now we want $\cot(t)+\dfrac1u=1$, or $\cot(t)+\cos(t)=1$. The problem now just reduces to finding the number solutions to that equation. However, this gives me an infinite number of solutions. Why is this wrong? Thanks!

share|cite|improve this question
up vote 5 down vote accepted

Let us continue your calculation at the point you reached $u=1+\sqrt{1+2^y}$. The calculation after that was not a good idea: introducing transcendental functions is unlikely to help in a problem of elementary number theory.

The solutions are $u=1\pm\sqrt{1+2^y}$. We want $\sqrt{1+2^y}$ to be an integer. So we want $1+2^y$ to be a perfect square, say $t^2$. We have arrived at the equation $1+2^y=t^2$ or equivalently $$2^y=t^2-1=(t-1)(t+1).$$ Now we can argue precisely as in the other solutions: $t-1$ and $t+1$ are each a power of $2$. That forces $t=\pm 3$. Since we are using $1\pm\sqrt{1+2^y}$, the choice $t=3$ is enough.

To finish, we have $u=1\pm 3$. So $u=-2$ or $u=4$. That gives $x=u+8=6$ or $x=12$. Each has $y=3$.

Remark: The OP mentions modular arithmetic. That cannot quite work. The reason is that the equation does have a solution. A pure congruential argument would show there are no solutions at all. We can use congruences to rule out certain types of solutions, but certainly not all.

share|cite|improve this answer
Thanks, I see how this is a better solution but why doesn't my solution work? cot(t)+cos(t)=1 has an infinite number of solutions. (I edited my question again, people keep editing the question and making mistakes). – Ovi May 25 '13 at 18:49
@Ovi: We do sort of have to worry about $1-\sqrt{1+2^y}$. It gives a solution. (Well, we don't have to worry about it if we note that the solutions of $2^y+1=t^2$ have $t=\pm 3$. But we do have to use one or the other.) Note that there are two $x$ that work in the original problem, but only one $y$. – André Nicolas May 25 '13 at 19:14
As to the $\cot t+\cos t =1$, there is nothing wrong with it. But you will be looking for solutions of this equation of very specific form. So it still leaves a problem to be solved. – André Nicolas May 25 '13 at 19:19
Could you elaborate on what specific form of solutions would I be looking for? And it is not the case that only some solutions of cot(t)+cos(t)=1 work, because according to wolframalpha all the solutions are wrong; they are not even integers: – Ovi May 25 '13 at 19:25
I should not have said the trig equation is OK. If I understand your triangles correctly, we have $\cos t=\frac{1}{u}\sqrt{u^2-1}$. But $\sqrt{u^2-1}$ is not $\cot t$ of the first triangle. – André Nicolas May 25 '13 at 19:39

Using your idea of substitution but without applying quadratic equations stuff:

$$u,y\in\Bbb Z\;,\;\;u(u-2)=2^y\iff u, u-2\;\;\text{are powers of}\;\;2, $$

and the only possibility I see is


share|cite|improve this answer


$(x-8)(x-10)=2^y \implies x-8=2^m$ and $x-10=2^k$. Only possible way such that both are powers of two and differ by $2$?


$2(2^{m-1}-2^{k-1})=2 \implies 2^{m-1}-1=1 \implies 2^{m-1}=2$, $m=2$ and $k=1$

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.