A Magic Property of Integers

Number Theory Level 2

Given positive integers $a, b, c,s$ .

Given $a^s+b^s=c^s$ .

Given $s\leq s_1$ for $a+b=c$ ,

$s\leq s_2$ for $a+b>c$ .

Find $s_1+s_2$ .

The answer is 3.

2 solutions

A Former Brilliant Member
Aug 20, 2020

$s_1$ :

$a^{s_1}+b^{s_1}=(a+b)^{s_1}$

From binomial theorem $s_1=0,1$ , but $s_1$ should be positive: $s_1=1$

$s_2$ :

From Fermat's last theorem $s_2\leq2$ . If $s_2=1$ , then $s=1$ , but if $a+b=c$ , then $a+b>c$ doesn't true. So $s_2=2$

$1+2=\boxed{3}$

In the first case, even if $s$ be allowed to assume the value $0$ , it won't be $0$ , since then the equation $a^0+b^0=(a+b)^0$ would be invalid. There are many theorems stated by Fermat. You should mention Fermat's last theorem .

A Former Brilliant Member - 9 months, 3 weeks ago

Yeah, but this is longer a bit :)

A Former Brilliant Member - 9 months, 3 weeks ago

Fermat fixed

A Former Brilliant Member - 9 months, 3 weeks ago

A Former Brilliant Member
Aug 20, 2020

By Fermat's last theorem , $s\leq 2$

For $a+b=c$ , $s$ is obviously equal to $1$

For $a+b>c,s\leq 2$

So, $s_1=1,s_2=2\implies s_1+s_2=\boxed 3$ .

A Magic Property of Integers

The answer is 3.

2 solutions

s 1 s_1 s 1 ​ :

s 2 s_2 s 2 ​ :

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$s_1$ :

$s_2$ :