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Solving equation

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anemone

MHB POTW Director
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Feb 14, 2012
3,681
Solve the equation $25^a+9^a+4^a=15^a+10^a+6^a$.
 
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kaliprasad

Well-known member
Mar 31, 2013
1,309
Solve the equation $25^a+9^a+4^a=15^a+10^a+6^a$.
Not a complete solution but

a cannot be >1

because $(25^a > (10^a + 15^a)$ and $9^a > 6^a$ so LHS is larger

if a = 1 LHS is larger

a = 0 is a solution and there may be more non integer solutions
 
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anemone

MHB POTW Director
Staff member
Feb 14, 2012
3,681
Solve the equation $25^a+9^a+4^a=15^a+10^a+6^a$.
Solution suggested by other:

$25^a+9^a+4^a=15^a+10^a+6^a$ is equivalent to $(2^a-3^a)^2+(3^a-5^a)^2+(5^a-2^a)^2=0$.

Obviously $a=0$ is the only answer to the problem and it happens when $2^a=3^a=5^a=1$.
 

Pranav

Well-known member
Nov 4, 2013
428
My solution:
The given equation can be written as:
$$(5^a)^2+(3^a)^2+(2^a)^2=5^a\cdot 3^a+5^a \cdot 2^a+3^a\cdot 2^a$$
Let $e=5^a$,$f=3^a$ and $g=2^a$. Then the given equation is:
$$e^2+f^2+g^2=ef+fg+ge$$
It is well known that $a^2+b^2+c^2 \geq ab+bc+ca$ and equality holds when $a=b=c$. Hence, in the given case: $e=f=g \Rightarrow 5^a=3^a=2^a$ and therefore, the only possible solution is $a=0$.