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

• Math100
In summary: 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.Let ##x=1835^{1910}+1986^{2061}+2134^{2209}\pmod{13}, y=1835^{1910}+1986^{2061}+2134^{
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} ##.

Math100 said:
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.

fresh_42 said:
Right. It took a moment to check the divisions.

And I find this funny.
How so?

Math100 said:
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.

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

Math100 said:
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.

Math100
Math100 said:
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
fresh_42 said:
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 ##?

Math100 said:
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.

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

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}?##

fresh_42 said:
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 ##.

Math100 said:
The smallest prime that divides ## 1835^{1910}+1986^{2061} ## is ## 7 ##.
Right. And why?

fresh_42 said:
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.

Math100 said:
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?

fresh_42 said:
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} ##?

Math100 said:
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.
Math100 said:
... 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.

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.

fresh_42 said:
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 ##.

Math100 said:
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.

Math100

## 1. What does it mean to prove something "mod 7"?

When we say "mod 7," we are referring to modular arithmetic, which is a mathematical system that deals with remainders. In this case, we are looking at the remainder when the expression 1835^1910 + 1986^2061 is divided by 7.

## 2. How do you prove a statement using modular arithmetic?

To prove a statement using modular arithmetic, we need to show that the given expression is congruent to 0 (or any other specific number) when divided by the chosen modulus (in this case, 7). This can be done by manipulating the expression algebraically and using properties of modular arithmetic.

## 3. What is the significance of the numbers 1835, 1910, 1986, and 2061 in this statement?

The numbers 1835 and 1986 are the bases of the exponents in the expression, while 1910 and 2061 are the exponents themselves. These numbers were likely chosen because they are large and can demonstrate the power of modular arithmetic in proving statements.

## 4. Can you provide an example of how to solve this statement using modular arithmetic?

Yes, we can solve this statement by using the property that (a+b)^n ≡ a^n + b^n (mod m). In this case, we have (1835^1910 + 1986^2061) ≡ (2^1910 + 1^2061) (mod 7). We can then simplify this to (2^1910 + 1) ≡ 0 (mod 7). Since 2^1910 ≡ 2^(3*636 + 2) ≡ (2^3)^636 * 2^2 ≡ 8^636 * 4 ≡ 1^636 * 4 ≡ 4 (mod 7), we have 4 + 1 ≡ 0 (mod 7), which is true.

## 5. Why is this statement important in the field of mathematics?

This statement is important because it demonstrates the power and usefulness of modular arithmetic in proving statements. It also shows the connection between different branches of mathematics, such as algebra and number theory. Additionally, this statement can be used in various real-world applications, such as cryptography and computer science.

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