Prove that ## 1835^{1910}+1986^{2061}\equiv 0\pmod {7} ##

[Math100]

Homework Statement

The three most recent appearances of Halley's comet were in the years $1835, 1910$, and $1986$; the next occurrence will be in $2061$. Prove that $1835^{1910}+1986^{2061}\equiv 0\pmod {7}$.

Relevant Equations

None.

Proof:

Observe that $1835\equiv 1\pmod {7}\implies 1835^{1910}\equiv 1\pmod {7}$.
Then $1986\equiv 5\pmod {7}$.
Applying the Fermat's theorem produces:
$5^{6}\equiv 1\pmod {7}$.
This means $1986^{2061}\equiv 5^{6\cdot 343+3}\pmod {7}\equiv 5^{3}\pmod {7}\equiv 6\pmod {7}$.
Thus $1835^{1910}+1986^{2061}\pmod {7}\equiv (1+6)\pmod {7}\equiv 0\pmod {7}$.
Therefore, $1835^{1910}+1986^{2061}\equiv 0\pmod {7}$.

 

[fresh_42]

Homework Statement:: The three most recent appearances of Halley's comet were in the years $1835, 1910$, and $1986$; the next occurrence will be in $2061$. Prove that $1835^{1910}+1986^{2061}\equiv 0\pmod {7}$.
Relevant Equations:: None.

Proof:

Observe that $1835\equiv 1\pmod {7}\implies 1835^{1910}\equiv 1\pmod {7}$.
Then $1986\equiv 5\pmod {7}$.
Applying the Fermat's theorem produces:
$5^{6}\equiv 1\pmod {7}$.
This means $1986^{2061}\equiv 5^{6\cdot 343+3}\pmod {7}\equiv 5^{3}\pmod {7}\equiv 6\pmod {7}$.
Thus $1835^{1910}+1986^{2061}\pmod {7}\equiv (1+6)\pmod {7}\equiv 0\pmod {7}$.
Therefore, $1835^{1910}+1986^{2061}\equiv 0\pmod {7}$.

Right. It took a moment to check the divisions.

And I find this funny.

 

[Math100]

Right. It took a moment to check the divisions.

And I find this funny.

How so?

 

[fresh_42]

How so?

Just so. No specific reason.

Who (and how) does someone find such an example? And what do we get if we added another appearance? It is just a little, cute incident. That's all.

 

[Math100]

I was thinking, that maybe we can observe the next occurrence, what do you think?

 

[fresh_42]

What do you mean by 'example'? Do you mean this problem?

Yes. It is an example of a congruence. I guess its randomness makes it fun for me.

The greatest conjectures began as a seemingly simple problem and turned out to establish entire areas in mathematics. $a^n+b^n=c^n$ looks innocent, doesn't it? Or "Any even number greater than 2 is the sum of two prime numbers." O.k. the problem above cannot really be compared to Fermat's last theorem or the Goldbach conjecture.

 

[fresh_42]

I was thinking, that maybe we can observe the next occurrence, what do you think?

The next appearances according to Wikipedia are $2134$ and $2209.$

Maybe you can find a small prime divisor in $1835^{1910}+1986^{2061}+2061^{2134}$ or $1835^{1910}+1986^{2061}+2061^{2134}+2134^{2209}.$

 

[Math100]

The next appearances according to Wikipedia are $2134$ and $2209.$

Maybe you can find a small prime divisor in $1835^{1910}+1986^{2061}+2061^{2134}$ or $1835^{1910}+1986^{2061}+2061^{2134}+2134^{2209}.$

Do you mean to find the smallest prime divisor? But how to find it? By considering modulo $2$?

 

[fresh_42]

Do you mean to find the smallest prime divisor? But how to find it? By considering modulo $2$?

You could use the same methods and try a few small primes. E.g. factorization of $2134$ and $2209$ might be a point to start with. The decision of whether these numbers are odd or even, i.e. modulo $2$ is very easy. But how many powers of $2$ are divisors?

It is only something to play with and use the tools you just earned.

More helpful would probably be to prove Fermat's little theorem.

 

[Math100]

The smallest prime divisor in $1835^{1910}+1986^{2061}+2061^{2134}$ is $2$.

 

[fresh_42]

We know $1835^{1910}+1986^{2061}\equiv 0\pmod{7}.$

What is the smallest prime that divides $1835^{1910}+1986^{2061}$?

And if we follow that pattern, what is the smallest prime that divides $1835^{1910}+1986^{2061}+2134^{2209}?$

 

[Math100]

We know $1835^{1910}+1986^{2061}\equiv 0\pmod{7}.$

What is the smallest prime that divides $1835^{1910}+1986^{2061}$?

And if we follow that pattern, what is the smallest prime that divides $1835^{1910}+1986^{2061}+2134^{2209}?$

The smallest prime that divides $1835^{1910}+1986^{2061}$ is $7$.

 

[fresh_42]

The smallest prime that divides $1835^{1910}+1986^{2061}$ is $7$.

Right. And why?

 

[Math100]

Right. And why?

Because $1835^{1910}\equiv 1\pmod {7}$ and $1986^{2061}\equiv 6\pmod {7}$. So their sum is $0\pmod {7}$. But I found out that $2134^{2209}\equiv 6\pmod {7}$ after finding $2134\equiv 6\pmod {7}$ and applying the Fermat's little theorem.

 

[fresh_42]

Because $1835^{1910}\equiv 1\pmod {7}$ and $1986^{2061}\equiv 6\pmod {7}$. So their sum is $0\pmod {7}$. But I found out that $2134^{2209}\equiv 6\pmod {7}$ after finding $2134\equiv 6\pmod {7}$ and applying the Fermat's little theorem.

This means that $7\,|\,\left(1835^{1910}+1986^{2061}\right)$ what we already knew. But why do we know that $2,3,5\,\nmid\,\left(1835^{1910}+1986^{2061}\right)?$

From $2134^{2209}\equiv 6\pmod {7}$ we get that $7\,\nmid\,\left(1835^{1910}+1986^{2061}+2134^{2209}\right)$ but what is the smallest prime divisor?

 

[Math100]

This means that $7\,|\,\left(1835^{1910}+1986^{2061}\right)$ what we already knew. But why do we know that $2,3,5\,\nmid\,\left(1835^{1910}+1986^{2061}\right)?$

From $2134^{2209}\equiv 6\pmod {7}$ we get that $7\,\nmid\,\left(1835^{1910}+1986^{2061}+2134^{2209}\right)$ but what is the smallest prime divisor?

I've tested the primes $2, 3, 5$ using Fermat's little theorem and they didn't work in $1835^{1910}+1986^{2061}$ and $1835^{1910}+1986^{2061}+2134^{2209}$. So following the pattern of $1835^{1910}+1986^{2061}\equiv 0\pmod {7}$, how can we find the smallest prime divisor of $1835^{1910}+1986^{2061}+2134^{2209}$?

 

[fresh_42]

I've tested the primes $2, 3, 5$ using Fermat's little theorem and they didn't work in $1835^{1910}+1986^{2061}$...

One of us has made a mistake.

... and $1835^{1910}+1986^{2061}+2134^{2209}$. So following the pattern of $1835^{1910}+1986^{2061}\equiv 0\pmod {7}$, how can we find the smallest prime divisor of $1835^{1910}+1986^{2061}+2134^{2209}$?

By trying. Or in this case: have a closer look.

 

[fresh_42]

Let
\begin{align*}
x&=1835^{1910}+1986^{2061}+2134^{2209}\pmod{13}\\
y&=1835^{1910}+1986^{2061}+2134^{2209}\pmod{17}\\
z&=1835^{1910}+1986^{2061}+2134^{2209}\pmod{19}
\end{align*}
Then $y-x=z\, , \,2y=3z\, , \,x+y+z=30.$ Prove that $5^3\,|\,xyz$ without using Fermat's little theorem.

 

[Math100]

Let
\begin{align*}
x&=1835^{1910}+1986^{2061}+2134^{2209}\pmod{13}\\
y&=1835^{1910}+1986^{2061}+2134^{2209}\pmod{17}\\
z&=1835^{1910}+1986^{2061}+2134^{2209}\pmod{19}
\end{align*}
Then $y-x=z\, , \,2y=3z\, , \,x+y+z=30.$ Prove that $5^3\,|\,xyz$ without using Fermat's little theorem.

Since $y-x=z$, it follows that $y=x+z, x=y-z, 3z=2y\implies z=\frac{2}{3}y, (y-z)+y+\frac{2}{3}y=30$.
Thus $2y=30\implies y=15$, so $x=5, z=10$.
Note that $xyz=5\cdot 15\cdot 10=750$.
This means $5^{3}\mid xyz\implies 125\mid 750$.
Therefore, $5^{3}\mid xyz$.

 

[fresh_42]

Since $y-x=z$, it follows that $y=x+z, x=y-z, 3z=2y\implies z=\frac{2}{3}y, (y-z)+y+\frac{2}{3}y=30$.
Thus $2y=30\implies y=15$, so $x=5, z=10$.
Note that $xyz=5\cdot 15\cdot 10=750$.
This means $5^{3}\mid xyz\implies 125\mid 750$.
Therefore, $5^{3}\mid xyz$.

Very nice. The congruences are (hopefully) right and I calculated them first, but they are only a distraction since the linear equations are sufficient.