Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

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

Solve $a^3-5a+7=3^b$ over the positive integer

I don't know how to solve such equation, please help me. Thanks

share|cite|improve this question
a=3k+1 (Using congruences) – Inceptio Mar 1 '13 at 13:49
$a=b=1$ is the only small solution. Tried up to $b=10$. – coffeemath Mar 1 '13 at 14:28
To add to Inceptio's answer: a must be 7 mod 9 for b=2, 16 mod 27 for b=3, 43 mod 81 for b=4, etc. None of these higher powers of 3 (9,27,81,...) that I tried have solutions. – coffeemath Mar 1 '13 at 14:31

Here is an idea which might or might not work:

Let $f(x)=x^3-5x+7 \,.$

$f(x)=0 \pmod 3$ has unique solution $x =1 \pmod 3$.

Moreover, $f'(x)= -5=1 \pmod 3$.

By the Lifting you can construct recursively $x_n \pmod {3^n}$ the uniquesolution to

$$f(x) =0 \pmod {3^n} \,.$$

Prove by induction that the smallest positive number $y_n$ in the class of $x_n \pmod {3^n}$ satisfies

$$y_n > 3^\frac{n}{3}$$

This shows that for $n=b$ all the solutions to $f(x)=3^b$ are not good.

share|cite|improve this answer
I'm not sure how one could make this induction work (it might, but controlling sizes in Hensel lifts can be rather delicate). – Mike Bennett Mar 1 '13 at 16:14

When $b=1$, we already know the result.

I can prove when $3\mid b$, there is no solution.

You may see

if we try $a = 3^{b/3}$, then

$a^3-5a+7 = 3^b-5\cdot3^{b/3}+7<3^b$, if $b\ge3$.

If we try $a = 3^{b/3}+1$, then

$a^3-5a+7 = 3^b+3\cdot 3^{2b/3}-2\cdot 3^{b/3}+3>3^b$, if $b\ge 3.$

Thus $3^{b/3}<a<3^{b/3}+1$, there is no $a$ satisfying this, when $3\mid b$.

share|cite|improve this answer
Are you sure about $3$ not dividing $39 = [3^{10/3}]$ ? – mercio Mar 1 '13 at 17:55
sorry I think I did something wrong. It is not true. – Yimin Mar 1 '13 at 17:56
you don't have to go very far to check that $[3^{b/3}]$ takes every possible modulus modulo $3$. $12 = \lfloor 3^{7/3} \rfloor$ – mercio Mar 1 '13 at 17:58
@mercio Yes, I found my mistake. So I can only prove the case $3\mid b$. already deleted that part. – Yimin Mar 1 '13 at 17:59

One way would be to find all the integral points on the two elliptic curves $$ y^2=a^3-5a+7 \; \mbox{ and } 3 y^2 = a^3 - 5a +7, $$ and to then look for values of $y$ that are powers of $3$. One could do this via, for instance, Magma. Appealing to Magma's IntegralPoints routine tells you that the first curve has only $(a,y)= (-2, \pm 3)$, while the second has $(a,y)=(1,\pm 1)$.

share|cite|improve this answer

if you start without positive integer restriction but real solution, you can express $b = \log_3(a^3-5a+7)$ and you want to find intersection: one solution is at [a=1, b=0] which is out of your target region, otherwise there are no solutions. This means, that there wouldn't be any integer slolutions, when there are no real solution at all.

share|cite|improve this answer
intersection with what? – N. S. Mar 1 '13 at 15:43
wait this is wrong, just pick any a larger than $5a-7+1$ and then you get a good $b$ – user58512 Mar 1 '13 at 15:43

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.