I have tried to solve this equation for $x, y >0$, $\log(x+y)=\log{x}\cdot \log{y}$ by writing it as this $x+y={x}^{\log{y}}$. But the last expression is complicated to me to get any solutions, I think it has no solution in integers !!!

My question here is : How do I show that for $x, y >0$, $\log(x+y)=\log{x} \cdot \log{y}$ has no solution in integers?

Note: $x, y$ are integer numbers.

Edit 01: I edited the question because I have a wrong meaning, I meant in integers.

Edit 02: I edited the question to show my key idea for the proof.

Attempt: This is just an Idea for the proof

What I think is : $\log(x+y)\leq\log(xy)=\log{x}+\log{y}$ which is less than $\log{x} \cdot \log{y}$ for $ x, y \geq 8$, hence: $2\leq \min(x,y)\leq 7$ , which it's easy to check.

Thank you for any help !!!!

  • $\begingroup$ sorry Daniel i meant in integers not in real numbers , i edited my question now $\endgroup$ – zeraoulia rafik Nov 12 '16 at 19:48
  • $\begingroup$ RHS IS logxlogy $\endgroup$ – zeraoulia rafik Nov 12 '16 at 21:03
  • $\begingroup$ how it's equivalent to x+y=x.y , i don't write :logx+logy =log(x.y) $\endgroup$ – zeraoulia rafik Nov 12 '16 at 21:05
  • $\begingroup$ Sorry, ignore my comment, that was nonsense... $\endgroup$ – Martin R Nov 12 '16 at 21:06
  • 1
    $\begingroup$ @dxiv I think Wolfram is wrong. I agree with you that it's a common mistake. For $\log_{e}$ there is $\ln$. $\endgroup$ – Michael Rozenberg Nov 13 '16 at 5:12

The following expansion is close to your idea and assuming $\log{y}$ is the natural logarithm.

Case 1. Let's assume $x=y$ then $\ln{2y}=\ln^2{y}$. By replacing $\ln{y}=z$ it's easy to check that $P(z)=z^2-z-\ln{2}=0$ has 2 solutions, the largest one $z_2$ is in between $(1,\frac{7}{4})$ and $\forall z > z_2: P(z)>0$. It's easy to check that $z_2 < \frac{7}{4}$ from $\ln{2} < \frac{21}{16}$.

Case 2. Let's assume $x>y$ or $x=qy+r, 0\leq r<y, q>0$ then $$\ln((q+2)y) > \ln((q+1)y+r)=\ln(qy+r)\ln{y} \geq \ln(qy)\ln{y}$$ or $$\ln(q+2) + \ln{y} > \ln{q}\ln{y} + \ln^2{y}$$ By replacing $\ln{y}=z$ $$P(z)=z^2+(\ln{q}-1)z-\ln(q+2)<0$$

Case 2.a $q=1$. This means $$P(z)=z^2-z-\ln{3}<0$$ another easy to check case that the largest solution $z_2$ of $P(z)$ is in between $(1,\frac{7}{4})$ (from $\ln{3} < \frac{21}{16}$) and $\forall z > z_2: P(z)>0$.

Case 2.b $q\geq 2$. We will check the roots of $P(z)$ $$D=\sqrt{(\ln{q}-1)^2+4\ln(q+2)}$$ and $$\sqrt{(\ln{q}-1)^2+4\ln{q}}<D<\sqrt{(\ln(q+2)-1)^2+4\ln(q+2)}$$ (including for $q=2$) or $$\ln{q}+1<D<\ln(q+2)+1$$ And the largest of 2 solution for $P(z)$ is $$z_2=\frac{-(\ln{n}-1)+D}{2}$$ or $$1<z_2<\ln{\sqrt{1+\frac{2}{q}}}+1<\frac{7}{4}$$ and in fact, with large $q$, is aggressively approaching $1$. In any case, the largest solution $z_2$ is in between $(1,\frac{7}{4})$ and $\forall z > z_2: P(z)>0$.

Note: regarding $$\ln{\sqrt{1+\frac{2}{q}}}+1<\frac{7}{4}$$ For $q\geq 2$ we have $\frac{3}{4} > \frac{1}{2} \ln{2} \geq \ln{\sqrt{1+\frac{2}{q}}} \Rightarrow \frac{7}{4} > \ln{\sqrt{1+\frac{2}{q}}} +1$

In all the cases, we have $\ln{y}=z_2 \in (1,\frac{7}{4})$ or $y \in \{3, 4, 5\}$. The remaining part is to check each $\ln(x+3)=\ln{x}\cdot \ln{3}$, $\ln(x+4)=\ln{x}\cdot \ln{4}$ and $\ln(x+5)=\ln{x}\cdot \ln{5}$ individually, but numerical approaches reveal no integer solutions.

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