# Relation between sum of two coprime numbers and sum of 37th power of the numbers

So you have two numbers $a$ and $b$ coprime. You then consider $\frac{a^{37} + b^{37}}{a + b}$ and $a + b$. When do these two values share common factors, and what are these common factors?

My Work

So the first term can be written as $\sum_{n=0}^{36}(-1)^{n}a^{n}b^{36-n} = a^{36}-a^{35}b^{1}+a^{34}b^{2}-\dots-ab^{35}+b^{36}$. The second term is still $a+b$. I've run this through a script on my HP-50g, and found that these two terms appear to be relatively prime iff 37 does not divide $a+b$. Conversely, if 37 divides $a+b$, then 11 is the only common prime factor of the two terms, and so gcd$\left(\frac{a^{37} + b^{37}}{a + b}, a+b\right)$ = 37.

Is there any way to prove this?

Note: $b\equiv -a\pmod {a+b}$ and:

\begin{align}\frac{a^{37}+b^{37}}{a+b}&=\sum_{n=0}^{36} (-1)^na^nb^{36-n} \\&\equiv \sum_{n=0}^{36}(-1)^n(-1)^{36-n}a^{36}\pmod{a+b}\\ &=a^{36}(1+1+\cdots +1)\\&=37a^{36}\pmod{a+b} \end{align}

Since $\gcd(a+b,a)=1$ you can conclude what?

More generally, if $u,v$ are distinct relatively prime integers and $d>0$ then:

$$\gcd\left(\frac{u^d-v^d}{u-v},u-v\right)=\gcd(d,u-v)$$

Here $u=a,v=-b,d=37$.

We are using three results here:

Theorem: If $m\equiv n\pmod d$ then $\gcd(m,d)=\gcd(n,d).$

Theorem: If $\gcd(m,n)=1$ then $\gcd(dm,n)=\gcd(d,n).$

Theorem: If $\gcd(m,n)=1$ then $\gcd(m^d,n)=1.$

Assume that $a+b = 0 \mod 37$, meaning $a+b= 37 k$ for $k \in \mathbb{N}$.

So we have $GCD(\frac{a^{37}+b^{37}}{a+b} , 37k)$

$a=37 k-b$, so we have $\frac{(37k-b)^{37}+b^{37}}{37 k} = \frac{b^{37}+\sum \limits_{j=0}^{37} \binom{37}{j} (37 k)^j (-b)^{37-j} }{37k}$

its obvious that when $j>1$ we have $37^2$ as multiple of some number.

and when $j=0$ we have $(-b)^{37} = -b^{37}$ and $-b^{37}+b^{37} =0$.

So we have $\frac{b^{37}+\sum \limits_{j=0}^{37} \binom{37}{j} (37 k)^j (-b)^{37-j} = 37^2 k m}{37 k} = 37 m$ for $m \in \mathbb{N}$.

Thus $GCD(\frac{a^37+b^37}{a+b},a+b) = 37$ if $a+b = 0 \mod 37$.

Use $$a\equiv -b\mod (a+b)$$ to derive $$\frac{a^{37}+b^{37}}{a+b}\equiv 37b^{36}\mod (a+b)$$

This means that every number $p$ dividing both $\frac{a^{37}+b^{37}}{a+b}$ and $a+b$ also divides $37b^{36}$. If some prime factor of $p$ would divide $b$, it would also divide $a$, contradicting the assumption $\gcd(a,b)=1$. Hence, $p$ must divide $37$, so you get $p=37$ as the only non-trivial possibility. So, the only non-trivial common factor is $37$. And we have this common factor if and only if $a+b$ is divisble by $37$.