5^n+3^n is increasing for all values of n; hence, if there's an integer solution, it must be unique. Indeed n=3 works and is, therefore, the only solution.
@@SALogics Surely there can only be one solution because both 3^n and 5^n are both monotonic and increasing. Though I can see that perhaps your solution is a general method that may work when trial and error doesn't give an obvious solution in a reasonable time.
@SALogics For N - yes, but for "5 rise to the power of (N/3)" this is generally not true. That is, the solution is found from a false assumption. But, fortunately, it was found and it was possible to prove that it is unique.
Методом подбора :n=3. Ваш способ решения правильный, но проще сделать таким образом 5^[n*2*(1/2)]+3^[n*2*(1/2)] =152 Замена:5^(п/2)=х 3^(п/2)=у х^2 +у ^2 =152
5^n+3^n is increasing for all values of n; hence, if there's an integer solution, it must be unique. Indeed n=3 works and is, therefore, the only solution.
Very nice! ❤
Thanks for the strategy
You are very welcome! ❤
By inspection n=3.
It is right
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yez n=3 and since the lhs is increasing and the rhs is constant there can only be one sokution.
@@SALogics Surely there can only be one solution because both 3^n and 5^n are both monotonic and increasing. Though I can see that perhaps your solution is a general method that may work when trial and error doesn't give an obvious solution in a reasonable time.
5^n + 3^n = 152
5^n + 3^n = 125 + 27
5^n + 3^n = 5^3 + 3^3
By comparing
n=3
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but where would you even get 125 + 27-
@@woskethebot 152 = 125+27
3
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n=3
(5*5*5)+(3*3*3)
125+27
152
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5^n+3^n=125+27=5^3+3^3 n=3
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5^n+3^n=152
=125+27=5^3+3^3
=>n=3
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Why not simply solve it by inspection as we do with cubic equations....
This is not allowed in olympiad! ❤
152=125+27 , 125+27=5^3+3^3 , 5^n-5^3=0 , 3^n-3^3=0 , --> n=3 , ..... ..... .....
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@@SALogics Thanks!
nice
Thanks! ❤
n=3, because 125+27=152.
Very nice! ❤
@SALogics Thank you!
X,y€ Z+ x+y < x^2 - xy + y^2 => WHY???? x=y=1 € Z+ , x+y > x^2 -xy +y^2 😢
x and y should'nt be same! ❤
Why are X and Y integers greater than zero? It's not obvious at all.
Because n is a positive integer!❤
@SALogics For N - yes, but for "5 rise to the power of (N/3)" this is generally not true. That is, the solution is found from a false assumption. But, fortunately, it was found and it was possible to prove that it is unique.
Методом подбора :n=3.
Ваш способ решения правильный, но проще сделать таким образом
5^[n*2*(1/2)]+3^[n*2*(1/2)] =152
Замена:5^(п/2)=х
3^(п/2)=у
х^2 +у ^2 =152
Very nice! ❤