6

So I have got that (a,b)=(1,0),(3,2) are solutions for the equations, and maybe the only one.

J. W. Tanner
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CryoDrakon
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  • If $b\ge 2$, $3^a\equiv 27\pmod{100}$, so $a\equiv 3\pmod {40}$. – ajotatxe Nov 23 '14 at 10:36
  • @ajotatxe: $a\equiv3\pmod{20}$ to be more accurate. – barak manos Nov 23 '14 at 10:42
  • For $a\le 5000,\ b\le 5000$, there are no more solutions. – Peter Nov 23 '14 at 11:11
  • Considering, how difficult it was to prove catalan's conjecture, it is probably very difficult to prove that there are no more solutions. – Peter Nov 23 '14 at 11:20
  • The least $j$ with $$5^j+2\equiv 0\ (\ mod\ 3^{11}\ )$$ is $70058$ – Peter Nov 23 '14 at 11:42
  • @Peter: Isn't it simple to show that if the "$$" part in "$25$" is divisible by $3$ then "$*27$" cannot be an integer power of $3$? – barak manos Nov 23 '14 at 12:06
  • I don't think so, but I looked at the continued fraction of $\frac{log(5)}{log(3)}$ and concluded that $a$ and $b$ have to be very, very large to give an extra solution. – Peter Nov 23 '14 at 12:10
  • In other words, it seems that $alog(3)-blog(5)$ cannot be close enoutgh to $0$, but I have to work out this idea. – Peter Nov 23 '14 at 12:11
  • Since $b$ is even, I guess these should be the only solutions of $3^a-x^2=2$. – Aravind Nov 23 '14 at 17:10
  • $x^2+2=(x+\sqrt{-2})(x-\sqrt{-2})=3^a$. Since $Z[\sqrt{-2}]$ is a UFD and $3=(1+\sqrt{-2})(1-\sqrt{-2})$, this should give a few possibilities, eg: $x+\sqrt{-2}=(1+\sqrt{-2})^a$ and $x-\sqrt{-2}=(1-\sqrt{-2})^a$. – Aravind Dec 06 '14 at 19:45

2 Answers2

4

We know that $b$ is even (since $2^b+2$ is divisible by 3). We also know that the only solution to $y^2+2=x^3$ is $y=5,x=3$. (Solving the diophantine equation $y^{2}=x^{3}-2$)

Thus it is sufficient to show that $a$ is divisible by 3. Suppose that $a \geq 2$. Since 9 divides $5^b+2$, we get that $b=6k+2=3m+2$. We have $25(125)^m+2=3^a$. We get $3^a$ is $27$ mod $31$ which forces $a$ to be $3$ mod $30$, in particular divisible by $3$.

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Aravind
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  • if ab=xy, and gcd(x,y)=gcd(a,b)=1 it doesn't mean that for example a has to be x or y, it can be a factor of one of those two or it can be the product of some factors of x and y. Like we have that $15\cdot 2=3\cdot 10$ – CryoDrakon Dec 06 '14 at 15:37
  • You are correct. I'll delete the answer for now. – Aravind Dec 06 '14 at 18:59
  • Also, the paper http://www.isibang.ac.in/~sury/x2+2=y%5En.pdf shows that this is the only solution to $x^2+2=y^n$. – Aravind Dec 08 '14 at 17:13
  • That would be it. Thank you :D – CryoDrakon Dec 08 '14 at 18:37
3

These are the only solutions modulo $$ 5^3 \cdot 601. $$ If you had a solution with $b \geq 3$ then we would require that $a \equiv 43$ modulo $100$. Modulo $601$ (there are lots of other choices), there are only $12$ choices for $5^b$. Solving $3^a \equiv 5^b+2 \mod{601}$ for each of these choices tells us that necessarily $a$ is congruent to $0, 1$ or $3$ modulo $75$, a contradiction.

Mike Bennett
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