MHB Prove that the expression is divisible by 26460

[anemone]

Problem:
Prove that $27195^8-10887^8+10152^8$ is divisible by $26460$.

Attempt:
I grouped the last two terms and manipulated them algebraically and came to the point where I suspect I might have taken the wrong path...here is the last step where I stopped and don't know how to proceed.

$\dfrac{27195^8-10887^8+10152^8}{26460}=\dfrac{3^5\cdot5^7\cdot7^{14}\cdot37^8-7013(2^6\cdot3^6\cdot47^2+3^2\cdot19^2\cdot191^2)(10152^4+10887^4)}{8}$

I'd like to ask, do you think this problem can be solved using only elementary methods?

Thanks in advance.

 

[Opalg]

Problem:
Prove that $27195^8-10887^8+10152^8$ is divisible by $26460$.

Attempt:
I grouped the last two terms and manipulated them algebraically and came to the point where I suspect I might have taken the wrong path...here is the last step where I stopped and don't know how to proceed.

$\dfrac{27195^8-10887^8+10152^8}{26460}=\dfrac{3^5\cdot5^7\cdot7^{14}\cdot37^8-7013(2^6\cdot3^6\cdot47^2+3^2\cdot19^2\cdot191^2)(10152^4+10887^4)}{8}$

I'd like to ask, do you think this problem can be solved using only elementary methods?

Thanks in advance.

The best way to make the calculation more manageable is to factorise $26460 = 2^2\cdot 3^3\cdot 5\cdot 7^2$. If you can separately show that $27195^8-10887^8+10152^8$ is divisible by each of the numbers $2^2$, $3^3$, $5$ and $7^2$, then the result will follow.

Take the factor 5, for example. Since $27195$ is a multiple of $5$, so is its eighth power. The other two terms are not multiples of $5$, but here you need to use your idea of grouping those two terms together. In fact, $a^8-b^8$ is a multiple of $a-b$. So $10887^8+10152^8$ is a multiple of $10887-10152 = 735$. That is a multiple of $5$. Putting those results together, you see that $27195^8-10887^8+10152^8$ is a multiple of $5$.

The exact same procedure shows that $27195^8-10887^8+10152^8$ is a multiple of $7^2$. You can use similar ideas to show that it is also a multiple of $3^3$ and of $2^2$. (To deal with $3^3$, notice that $a^8-b^8$ is also a multiple of $a+b$.)

 

[anemone]

The best way to make the calculation more manageable is to factorise $26460 = 2^2\cdot 3^3\cdot 5\cdot 7^2$. If you can separately show that $27195^8-10887^8+10152^8$ is divisible by each of the numbers $2^2$, $3^3$, $5$ and $7^2$, then the result will follow.

Take the factor 5, for example. Since $27195$ is a multiple of $5$, so is its eighth power. The other two terms are not multiples of $5$, but here you need to use your idea of grouping those two terms together. In fact, $a^8-b^8$ is a multiple of $a-b$. So $10887^8+10152^8$ is a multiple of $10887-10152 = 735$. That is a multiple of $5$. Putting those results together, you see that $27195^8-10887^8+10152^8$ is a multiple of $5$.

The exact same procedure shows that $27195^8-10887^8+10152^8$ is a multiple of $7^2$. You can use similar ideas to show that it is also a multiple of $3^3$ and of $2^2$. (To deal with $3^3$, notice that $a^8-b^8$ is also a multiple of $a+b$.)

Awesome! I finally understand it now!

Thanks for your help, Opalg!

 

[Nono713]

Opalg's answer is very good! A more "heavy machinery" method could also go as follows:

$$26460 = 2^2 \cdot 3^3 \cdot 5^1 \cdot 7^2$$
And we can do the following reductions with a couple modulo operations:

$$27195^8 - 10887^8 + 10152^8 \equiv 3^8 - 3^8 + 0^8 \equiv 0 \pmod{2^2}$$

$$27195^8 - 10887^8 + 10152^8 \equiv 6^8 - 6^8 + 0^8 \equiv 0 \pmod{3^3}$$

$$27195^8 - 10887^8 + 10152^8 \equiv 0^8 - 2^8 + 2^8 \equiv 0 \pmod{5^1}$$

$$27195^8 - 10887^8 + 10152^8 \equiv 0^8 - 9^8 + 9^8 \equiv 0 \pmod{7^2}$$
And invoking the CRT (since $2^2$, $3^3$, $5^1$, $7^2$ are pairwise coprime):

$$\mathbb{Z}_{26460} = \mathbb{Z}_{2^2} \times \mathbb{Z}_{3^3} \times \mathbb{Z}_{5^1} \times \mathbb{Z}_{7^2}$$
$$\therefore$$
$$27195^8 - 10887^8 + 10152^8 \equiv 0 \pmod{26460}$$
This also shows that the exponent (8) is in fact irrelevant, and could be anything.

EDIT: fixed, see ILikeSerena's post below.

 

[I like Serena]

Opalg's answer is very good! A more "heavy machinery" method could also go as follows:

$$26460 = 2^2 \cdot 3^3 \cdot 5^1 \cdot 7^2$$
And we can do the following reductions with a couple modulo operations:

$$27195^8 - 10887^8 + 10152^8 \equiv 0^8 - 0^8 + 0^8 \equiv 0 \pmod{2}$$

$$27195^8 - 10887^8 + 10152^8 \equiv 0^8 - 0^8 + 0^8 \equiv 0 \pmod{3}$$

$$27195^8 - 10887^8 + 10152^8 \equiv 0^8 - 2^8 + 2^8 \equiv 0 \pmod{5}$$

$$27195^8 - 10887^8 + 10152^8 \equiv 0^8 - 2^8 + 2^8 \equiv 0 \pmod{7}$$
And invoking the CRT (since 2, 3, 5, 7 are pairwise coprime):

$$\mathbb{Z}_{26460} = \mathbb{Z}_{2^2} \times \mathbb{Z}_{3^3} \times \mathbb{Z}_{5^1} \times \mathbb{Z}_{7^2}$$
$$\therefore$$
$$27195^8 - 10887^8 + 10152^8 \equiv 0 \pmod{26460}$$
This also shows that the exponent (8) is in fact irrelevant, and could be anything.

I'm afraid that the number $2 \cdot 3 \cdot 5 \cdot 7$ is also zero mod 2, mod 3, mod 5, and mod 7.
But it is not divisible by 26460.

For CRT you need to show the mod relation for $2^2$, $3^3$, $5^1$, and $7^2$.

For instance for $3^3$, we need:

$\hspace{0.5 in}27195^8 - 10887^8 + 10152^8 \equiv 0^8 - 0^8 + 0^8 \equiv 0 \pmod{3^3}$

Still quite doable.

 

[Nono713]

I'm afraid that the number $2 \cdot 3 \cdot 5 \cdot 7$ is also zero mod 2, mod 3, mod 5, and mod 7.
But it is not divisible by 26460.

For CRT you need to show the mod relation for $2^2$, $3^3$, $5^1$, and $7^2$.

For instance for $3^3$, we need:

$\hspace{0.5 in}27195^8 - 10887^8 + 10152^8 \equiv 0^8 - 0^8 + 0^8 \equiv 0 \pmod{3^3}$

Still quite doable.

True, my mistake. Though it's not too difficult to check, just do the same operations but using 4, 27 and 49... (which still work out, it's clear the integers in the problem were carefully chosen to cancel each other out)