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

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

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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.

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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 $3\mid b$.

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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 $(a,y)=(1,\pm 1)$.

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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.

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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