Find solution following the system of equations!

Question

$abcde-a=357^{400}$

$abcde-b=359^{410}$

$abcde-c=361^{420}$

$abcde-d=363^{430}$

$abcde-e=365^{440}$

($a,b,c,d,e$ are natural numbers)

I don't have any idea.

I just tried this, $$a(bcde-1)=357^{400}$$

$$b(acde-1)=359^{410}$$

$$c(abde-1)=361^{420}$$

$$d(abce-1)=363^{430}$$

$$e(abcd-1)=365^{440}$$

so $$b=359^k $$ , for some $k$ since $359$ is prime number.

But this is of no service to find solution.

This is so hard for me.

anyone know this?

pleas help me.. thanks.

Answer 1

It is rather simple to prove that there is no integer solution.

Suppose there is a solution for integers $a$, $b$, $c$, $d$, and $e$.

From the first equation we have that $a$ divides $a (b c d e - 1) = 357^{400}$ and hence needs to be odd.

Likewise, the other integers have to be odd as well.

But then all the equations are of the form:

odd - odd = odd

Hence, there is no integer solution.

Answer 2

I have a solution that uses nothing special, just high school mathematics.
$$a= abcde-357^{400}$$ $$b= abcde-359^{410}$$ $$c= abcde-361^{420}$$ $$d= abcde-363^{430}$$ $$e= abcde-365^{440}$$

Multiply these. Then you obtain a quintic equation for $x=abcde$: $x=(x-357^{400})(x-359^{410})(x-361^{420})(x-363^{430})(x-365^{440})$

In particular $x$ divides $357^{400}\cdot 359^{410}\cdot 361^{420} \cdot 363^{430} \cdot 365^{440}$

Clearly $x\neq 0$. On the right hand side every term is an integer, so the absolute value of any difference is at least 1. But those numbers that we subtract are very far apart (they are in increasing order). It is easy to see that $|(x-363^{430})(x-365^{440})|\geq \frac{1}{2}\cdot 365^{440}$, and $(x-359^{410})(x-361^{420})\geq \frac{1}{2}\cdot 361^{420}$. So the absolute value of the product on the right is at least $100\cdot 365^{440}$. But then $x\geq 100\cdot 365^{440}$. Looking at the right hand side again, this means that every differnce is at least $99\cdot 365^{440}$. But then the product is at least $(99\cdot 365^{440})^5$, which is much bigger than $357^{400}\cdot 359^{410}\cdot 361^{420} \cdot 363^{430} \cdot 365^{440}$, so $x> 357^{400}\cdot 359^{410}\cdot 361^{420} \cdot 363^{430} \cdot 365^{440}$. But then $x$ cannot divide $357^{400}\cdot 359^{410}\cdot 361^{420} \cdot 363^{430} \cdot 365^{440}$, a contradiction. So there is no solution.

Answer 3

Some thoughts (but I don't know if they are valid for middle school childs) :

1. of the five numbers $357$, $359$, $361$, $363$, $365$, two are very interesting : $359$, which is prime, and $361$ which is a power of a prime ($19^2$).

2. So what one can do is remark $b(acde-1)=359^{410}$, so $b$ divides $359^{410}$, and $b$ is a power of $359$, say $359^\beta$.

3. Equally, $c$ divides $19^{840}$, therefore $c=19^\gamma$.

4. Now you can write $abcde=p=359^{410}-359^\beta=19^{840}-19^\gamma$.

5. As $u^n-v^n=(u-v)\sum_{k=0}^{n-1}u^kv^{n-1-k}$, you can see that $18$ divides $19^{840}-19^\gamma$, and $358=2.179$ divides $359^{410}-359^\beta$.

6. Also, you have $19$ and $359$ primes, hence co-primes, so $19$ must divide $359^{410-\beta}-1$ and $359$ must divide $19^{840-\gamma}-1$.

7. And so on... Maybe this can get you to the solution(s), or to a proof that there are none :-)