# Proving that $2$ is the only real solution of $3^x+4^x=5^x$

I would like to prove that the equation $3^x+4^x=5^x$ has only one real solution ($x=2$)

I tried to study the function $f(x)=5^x-4^x-3^x$ (in order to use the intermediate value theorem) but I am not able to find the sign of $f'(x)= \ln(5)\times5^x-\ln(4)\times4^x-\ln(3)\times3^x$ and I can't see any other method to solve this exercise...

• A solution $x > 2$ is not possible by Fermat's Last Theorem. – user9413 Sep 4 '11 at 13:28
• @Chandrasekhar : A rational solution $x > 2$ is not possible by FTL, but it says nothing about irrational solutions. – Gunnar Þór Magnússon Sep 4 '11 at 13:29
• Maybe it would be easier to write the equation as $\left(\frac 35\right)^x +\left(\frac 45\right)^x=1$. – Davide Giraudo Sep 4 '11 at 13:51
• Write $x = 2n + r$ for $n\in\mathbb{Z}$ and $r\in[0,2).$ Then if $x>2,$ $$5^x = (4^2 + 3^2)^n 5^r = 4^{2n}5^r + 3^{2n}5^r + 5^r\displaystyle\sum_{i = 1}^{n-1} \binom {n} {i} 4^{2i}3^{2(n-1)} > 4^{2n + r} + 3^{2n+ r} = 4^x + 3^x.$$ – jspecter Sep 4 '11 at 13:56
• And if $x \in \mathbb{Q}_{<0}$ the lhs lies in $\mathbb{Z}_{(5)}$ whereas the rhs does not. – jspecter Sep 4 '11 at 14:03

One direct method is to divide directly by $5^x$ and get $1=(3/5)^x+(4/5)^x$. From here it is clear that the RHS is strictly decreasing, and there is a unique solution. Almost all exponential equations can be treated this way, by transforming them to

• one increasing function equal to one decreasing function

• one increasing/decreasing function equal to a constant.

• Thanks for sharing this method. I am wondering how could we prove that (a) the function is continuously decreasing (other than by piloting or inspection) and (b) there is a unique solution. Thanks. – NoChance Jul 2 '19 at 15:33
• I guess it is possible to prove that the function is strictly increasing without calculus, just using the basic properties of the exponential function $x \mapsto a^x$ for $a>1$ or $a<1$. This will prove that there is at most one solution. The uniqueness comes from the fact that in this case we can identify the solution $x=2$. – Beni Bogosel Jul 2 '19 at 20:56
• Thank you for your answer. – NoChance Jul 2 '19 at 20:58

If we insert the known solution we can write $$5^{2+x} = 4^{2+x} + 3^{2+x}$$ asking, whether there might another solution exist besides $x=0$ . Then we can rewrite, putting the $5^x$ to the rhs:
$$5^2 = 4^2\cdot 0.8^x + 3^2\cdot 0.6^x$$ Then if the exponents $x$ on the rhs are zero, we have the known solution. But if $x$ increases over zero, then the values of both summands decrease simultaneously, thus the equality can no more hold.
The analogue occurs for decreasing $x$: both summands increase over their squares simultaneously, so there is no other solution possible. QED.

• Best possible answer i think. – mick Sep 29 '12 at 19:18
• +1 Marvelous answer...but kind of related to Beni's answer too – imranfat Mar 2 '17 at 4:10