1. ## Product of Primes

Let $q_i, p$ be distinct odd primes $\alpha,k>1$ positive integer

Show that both of these identities cannot be true simultanously:

$
\prod_{i=1}^{k}q_i = p^\alpha+2
$

$
\prod_{i=1}^{k}(q_i-1) = p^\alpha+2-p^{\alpha-1}
$

I have proved it for k=2 but have no idea how to prove it for higher k

2. Originally Posted by wauwau
Let $q_i, p$ be distinct odd primes $\alpha,k>1$ positive integer

Show that both of these identities cannot be true simultanously:

$
\prod_{i=1}^{k}q_i = p^\alpha+2
$

$
\prod_{i=1}^{k}(q_i-1) = p_i^\alpha+2-p^{\alpha-1}
$

I have proved it for k=2 but have no idea how to prove it for higher k
Try "proof by induction". But you have " $p_i$ in your second formula and you have not defined it.

3. I have corrected the error.

But since the right sides of the equation change completely when you increase the k there is no way of attacking this with induction - from my point of view!!

4. the euler function give me some idea, but I didn't solve it completely yet.
You can try, I wish you success!

5. If we apply the euler function to the product of $q_i$ , and combined with the second equality, we have:
$2^k-1=p^{\alpha-1}$ (&)
thus $\alpha$ is not greater than k.
combined with the first equality, gives p is less than the greatest prime among $q_i$.
from (&), we see that
$\prod_{i=1}^{k}q_i = p^\alpha+2 = p(2^k-1)+2 < 2^kp$
that is impossible since $q_i,p$ are distinct odd prime!

is it correct?

6. Originally Posted by Shanks
If we apply the euler function to the product of $q_i$ , and combined with the second equality, we have:
$2^k-1=p^{\alpha-1}$ (&)
thus $\alpha$ is not greater than k.
combined with the first equality, gives p is less than the greatest prime among $q_i$.
from (&), we see that
$\prod_{i=1}^{k}q_i = p^\alpha+2 = p(2^k-1)+2 < 2^kp$
that is impossible since $q_i,p$ are distinct odd prime!

is it correct?
Can you explain, where the $2^k-1$ comes from?

7. apply the euler phi function to the product of q_i.
$2^k-1$ is the number of numbers which satisfy
(1)not greater than the product of q_i.
(2)not relatively prime to the product of q_i.

8. Originally Posted by Shanks
apply the euler phi function to the product of q_i.
$2^k-1$ is the number of numbers which satisfy
(1)not greater than the product of q_i.
(2)not relatively prime to the product of q_i.
How so? You're saying that $q_1\dots q_k - \varphi(q_1\dots q_k) = 2^k-1$? If so, this is clearly false.

9. Thanks，I know where I go wrong!