Matematické Fórum

stuart clark · 24. 11. 2011 04:28

solve for $x$ and $y$

$2^x+3^x+4^x=y^2$

where $x\;y\in\mathbb{Z}$

eldest · 24. 11. 2011 08:53

x = 1
y = 3

I get it very easily I said to myself what if X = 1 ?

$2 + 3 + 4 = 9$

What is 9? It is $3^{2}$

vanok · 24. 11. 2011 11:06 — Editoval vanok (24. 11. 2011 11:48)

Hi ↑ eldest:,

Yes, completely.
But it remains to show that it is the only solution.
Thus let us suppose that $x>1$ .
The left expression mod 4 is $3^x$
We deduct from it that $x$ can to be one squared that if it is odd.
Let us look at the same modulo expression 3. For $x$ odd, the expression is congru in 2 mod 3.
What is impossible.
We notice that for $x> 1$ does not solution.
For $x$ negatives, the sum is strictly lower than 1, thus not give a solution. Safe for $x=-1$ , but which does not also suit.
$x=0$ does not solution.

Conclusion: x=1 is the only solution.

Sincerely Vanok

Pavel · 24. 11. 2011 11:32

↑ stuart clark:

Autor skryl část textu. Zobrazit/skrýt!

Pavel · 24. 11. 2011 11:35

↑ vanok:

It is the same idea as mine. But it is not true that for $x$ negatives the sum is strictly less than 1, try $x=-1$ .

vanok · 24. 11. 2011 11:47

↑ Pavel:,
Yes, 2 solutions have the same ideas.
I forgot to copy out a sentence: Safe for $x=-1$ , but which does not also suit.
I add it ini my text...thank you
Sincerely Vanok

stuart clark · 25. 11. 2011 17:08

thanks pavel, vanok and eldest

Matematické Fórum

#1 24. 11. 2011 04:28

exponential equation

#2 24. 11. 2011 08:53

Re: exponential equation

#3 24. 11. 2011 11:06 — Editoval vanok (24. 11. 2011 11:48)

Re: exponential equation

#4 24. 11. 2011 11:32

Re: exponential equation

#5 24. 11. 2011 11:35

Re: exponential equation

#6 24. 11. 2011 11:47

Re: exponential equation

#7 25. 11. 2011 17:08

Re: exponential equation

Zápatí