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.