$3^x+4^x+5^x=6^x(R:x=3)$
I try but I can't finish
$3^x+2^x~2^x+5^x=3^x.2^x$
$3^x\cdot2^x-3^x-2^x\cdot2^x=5^x$
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3I think there is nothing better than to guess the solution (or if this is not possible to apply numerical methods) – Peter Jan 16 '23 at 13:59
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2Are you assuming that $x$ is an integer? – Dietrich Burde Jan 16 '23 at 14:12
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@DietrichBurde x is an integer – peta arantes Jan 16 '23 at 18:13
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1A recent suggestion on meta concerning the tag (math-school): https://math.meta.stackexchange.com/a/35471. (I thought it might be reasonable to let you know - as far as I can tell, you're the creator of this tag.) – Martin Sleziak Jan 31 '23 at 15:16
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Comment:
In addition to Dietrich's algebraic solution, you may also apply this elementary reasoning:
$4^x+5^x=6^x-3^x=3^x(2^x-1)=3 y$
this relation deduces that left hand side must also be a multiple of 3. it can be seen that only with odd x this is possible:
$x=1\Rightarrow 4^1+5^1=9= 3\times 3$
$x=3\Rightarrow 4^3+5^3=64+125=189=3\times 63$
etc. You can check theses powers and find that :
$4^3+5^3=189=27\times 7=3^3(3^2-1)$
and conclude that $x=3$
Or:
$3^3+4^3+5^3=6^3$
sirous
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"You can check ...and conclude that $x=3$". I don't see this. You have $x$ odd as restriction so far. So $x=\pm 1,\pm 3,\pm 5, \cdots $. How do you conclude that $x=3$? – Dietrich Burde Jan 16 '23 at 19:11