be non-negative real numbers.Prove that :![\[ \sqrt{x^4+y^4 } +(2+\sqrt{2})xy \geq x^2+y^2 \]](http://latex.artofproblemsolving.com/6/2/3/623d50afffd81b2783ff98568791164b53747567.png)
?
be real numbers such that 
Prove that
Let 
Prove that
Prove that
Prove that
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Find all pairs 
of positive integers such that ![\[ k!=(2^n-1)(2^n-2)(2^n-4)\cdots(2^n-2^{n-1}). \]](//latex.artofproblemsolving.com/3/5/3/353bbb15bac205053c26208af8cac7c32a296f55.png)
Proposed by Gabriel Chicas Reyes, El Salvador
of positive integers such that ![\[ k!=(2^n-1)(2^n-2)(2^n-4)\cdots(2^n-2^{n-1}). \]](http://latex.artofproblemsolving.com/3/5/3/353bbb15bac205053c26208af8cac7c32a296f55.png)
Proposed by Gabriel Chicas Reyes, El SalvadorThis post has been edited 4 times. Last edited by v_Enhance, Jul 18, 2019, 6:26 PM
Reason: add $2^n-4$ term
Reason: add $2^n-4$ term
is more important than a missed comma. This can be confused to 
'
,
.
.
'
by the LTE lemma, so 
.
,
.
)
and 
for left and right side respectively.
.
.
,
.
,
,
,
.
are the only solutions.
When 
i.e. 
as 
.
to get 
and 
,
.
,
,
which is trivially false for 
as there are 
terms on both sides, and the other case is very similar, though a little more complicated-the answers will be 
for 
respectively).
.![\[ \nu_2 (k!) = \nu_2 \left( \prod_{i=0}^{n-1} (2^n - 2^i) \right) = \frac{(n(n-1)}{2} . \]](http://latex.artofproblemsolving.com/7/2/3/7234d348576e0e3c8ff02ea030007fe79248badd.png)
Thus 
.![\[ \prod_{i=0}^{n-1} (2^n - 2^i) > 2^{n^2}. \]](http://latex.artofproblemsolving.com/b/8/d/b8d2a4745f458fcc08dcbda034a5c8109da4205f.png)
It is easy to see that these two facts are impossible for sufficiently large 
.
will suffice since
Then we induct up:![\[ \left(\frac{n(n+1)}{2}\right)! > 2^{n^2} \cdot 6^{n} > 2^{(n+1)^2} \]](http://latex.artofproblemsolving.com/3/4/d/34d88799270ef7ae5bc81222228fcedf36de1c5e.png)
for all 
.
.
are solutions, but when 
,
has no solution, note that the right handside is divisible by 31, so then 
,
But we also have:
Then we must have 
So we must have :
This holds only for n<7. We can check 
by hand.
,
= n(n-1)/2 thus 
, now RHS<
thus k!<
>
> 
if 
Now, from LTE, we have that 
.
By checking manually we see that 
and 
,
and 
to see they do not have an integer solution for 
.
.
,
divisible by 
contributes at least one factor of 
.
.
is 
and 
,
,
.
for all 
,
of both sides we have,
Thus, 
.
if and only if 
.
.
Thus, as a result of our previous bound we have,
which implies,![\[\frac{n}{6}-1<\lfloor \frac{n}{6} \rfloor\le \nu_3((\lfloor \frac{n}{2} \rfloor)!) \le \frac{\lfloor \frac{n}{2} \rfloor}{n-2}\]](http://latex.artofproblemsolving.com/b/b/a/bba35b874f0f61eece62e9a7f540737caed7d8a1.png)
from which it is not hard to see that we require 
.
we see that 
and 
work yielding 
and 
respectively and others don't 
,
.
as well. So, 29 is a fraction of some expression of the form 
which is clearly impossible for any 
which completes the check.
be the sum of digits in the base 
expansion of 
.![\[ \nu_2(Q) = \frac{n(n-1)}{2}\]](http://latex.artofproblemsolving.com/a/6/7/a6720ea5962b7d0ea0948953ee2d25a289d4caed.png)
That shows 
.![\[ \nu_p(k!) \leq \lfloor \frac{2^n}{p-1}\rfloor \leq \nu_p(Q)\]](http://latex.artofproblemsolving.com/1/3/3/133030946003860c1aaa3ae201d8d9b209cd8c00.png)
Clearly for all primes the 
makes no diffrence. now the equality should hold for that 
we see that only 
work
and 
.
to balance out the other prime factors. Specifically, in addition to examining 
.![\[\nu_2(k!) = \sum_{i = 0}^{n - 1} \nu_2(2^n - 2^i) = 1 + 2 + \dots + (n - 1) = \frac{n(n - 1)}{2}\]](http://latex.artofproblemsolving.com/7/2/2/722ef53213b209dda2ef9fd441696edfffeede2b.png)
and ![\[\nu_3(k!) = \nu_3\left(\prod_{i = 0}^{n - 1} (2^n - 2^i)\right) = \sum_{i = 1}^n \nu_3(2^i - 1) = \sum_{i = 1}^{\lfloor n/2 \rfloor} \nu_3(4^i - 1) = \left\lfloor \frac n2 \right\rfloor + \sum_{i = 1}^{\lfloor n/2 \rfloor} \nu_3(i)\]](http://latex.artofproblemsolving.com/1/5/b/15b3c78038390215e3ca158812562630e0a22d66.png)
![\[ = \left\lfloor \frac n2 \right\rfloor + \nu_3\left(\left\lfloor \frac n2 \right\rfloor!\right)\]](http://latex.artofproblemsolving.com/c/e/a/cea539d00d3c035e331b12a5765f3a90ae6bf928.png)
by LTE and the fact that 
is 
when 
,![\[\nu_3\left(\left(\left\lfloor \frac n2 \right\rfloor + 1\right) \cdot \left(\left\lfloor \frac n2 \right\rfloor + 2\right) \cdots k\right) = \left\lfloor \frac n2 \right\rfloor.\]](http://latex.artofproblemsolving.com/e/9/c/e9cac8a489be49fa9801c86ff47de09d73000ebd.png)
At least 
terms in the product on the left side are divisible by 
,![\[\left\lfloor \frac n2 \right\rfloor \geq \left\lfloor \frac{k - \lfloor n/2 \rfloor}{3} \right\rfloor \geq \frac{k - \lfloor n/2 \rfloor}{3} - 1 \implies k \leq 4 \left\lfloor \frac n2 \right\rfloor + 3 \leq 2n + 3.\]](http://latex.artofproblemsolving.com/8/3/c/83ca8e2cf43596a8c68d6d62d174d57d2cb4d613.png)
But from before we have ![\[\frac{n(n - 1)}{2} = \nu_2(k!) = k - s_2(k) < k,\]](http://latex.artofproblemsolving.com/6/3/7/63736929ca5258a66bede6fb9edde5a5519be0ba.png)
so 
.
,
.
and 
work as desired. 
then no such 
.
if 
and we can check that 
if 
.
which is equivalent to the lemma. ////
.
Now we state another lemma.
for 
.
iff 
.
Since 
it follows that 
.
So from 
and 
it follows that 
as claimed.
we can manually check and find that only 
only, it is trivial to verify that both of these work.
,
which is just 
,
,
.
.
we have 
,
,
is true for most 
the inequality 
is true, which is just combining the three highest terms with the three lowest. This is just obviously true by computation and the fact that the function is increasing.
![\[\frac{n(n-1)}2=\nu_2\left(\prod_{i=0}^{n-1}(2^n-2^i)\right)=\nu_2(k!)\le k\]](http://latex.artofproblemsolving.com/0/3/7/037b148c6003a97b31b0b09ec7a7670951ea423c.png)
where the last step follows from evaluating Legendre's formula as a geometric series.![\[\frac{k}3-1<\nu_3(k!)=\nu_3\left(\prod_{i=0}^{n-1}(2^n-2^i)\right)=\nu_3\left(\prod_{i=1}^{n}(2^i-1)\right).\]](http://latex.artofproblemsolving.com/6/0/9/6097e98727e3b3950cdac8dc73e482f91e453a47.png)
Now 
if 
is odd and 
if ![\[\frac{k}3-1<\nu_3\left(\prod_{i=1}^{n}(2^i-1)\right)\le\frac{n}2+\sum_{i=1}^{\left\lfloor\frac{n}2\right\rfloor}\nu_3(2i)\le\frac{n}2+\nu_3(n!)<n\]](http://latex.artofproblemsolving.com/a/2/5/a25e3d650555c80829594f60c3797e3a18ee916f.png)
where the last step again follows from evaluating Legendre's formula as a geometric series.
,
.
.
and 
.
and 
.
,![\[v_5(k!) \geq v_7(k!)\]](http://latex.artofproblemsolving.com/f/2/b/f2beb3f8c83a8e34c2b44ba6d3ba6f0cbb3d534f.png)
This is trivial by Legendre's. Now, let![\[P(n) = \prod_{i = 0}^{n - 1} (2^n- 2^i)\]](http://latex.artofproblemsolving.com/4/8/8/488abc99a72779408a434c9d239d6b8414fa8115.png)
Thus, if there were a solution pair 
we would need![\[v_5(P(n)) \geq v_7(P(n))\]](http://latex.artofproblemsolving.com/d/0/7/d0798640f22222ed0a3f244e9766cdb3ef809216.png)
Now, note that the order of 
is 
respectively. So, by LTE![\[v_5(2^{4k} - 1) = v_5(k) + 1\]](http://latex.artofproblemsolving.com/3/a/6/3a6c3ed57b749485437b8ca1efc8689460834ea8.png)
![\[v_7(2^{3k} - 1) = v_7(k) + 1\]](http://latex.artofproblemsolving.com/e/1/e/e1ee9b93c860552121438fe02782428da1eebc17.png)
Thus, we in fact have![\[v_5(P(n)) = \left\lfloor \frac{n}{4} \right\rfloor + \left\lfloor \frac{n}{4 \cdot 5} \right \rfloor + \cdots\]](http://latex.artofproblemsolving.com/5/7/b/57b285a50087b6d9edc4b9280654950eece4f1e5.png)
![\[v_7(P(n)) = \left\lfloor \frac{n}{3} \right\rfloor + \left\lfloor \frac{n}{3 \cdot 7} \right \rfloor + \cdots\]](http://latex.artofproblemsolving.com/9/3/b/93b9dad4af0ee0c3a133b7710c9169be095ed9c5.png)
Now, notice that![\[v_5(P(n)) = \left\lfloor \frac{n}{4} \right\rfloor + \left\lfloor \frac{n}{4 \cdot 5} \right\rfloor + \cdots < \frac{n}{4} + \frac{n}{4 \cdot 5} + \cdots = \frac{5n}{16}\]](http://latex.artofproblemsolving.com/6/e/e/6ee765ae81d6a8045a7dec3ea7a079830554b92b.png)
Furthermore, if 
,![\[v_7(P(n)) > \frac{n}{3} r + \cdots + \frac{n}{3 \cdot 7^m} - (m + 1) \geq \frac{n}{18}(\frac{7^{m + 1} - 1}{7^m}) - (m + 1)\]](http://latex.artofproblemsolving.com/7/2/b/72b0c6b9676153b313c9b182bee91dafd4083535.png)
Assume 
,![\[v_7(P(n)) > \frac{n}{3} r + \cdots + \frac{n}{3 \cdot 7^m} - (m + 1) \geq \frac{n}{18}(\frac{7^{m + 1} - 1}{7^m}) - (m + 1) > \frac{8n}{21} - m - 1\]](http://latex.artofproblemsolving.com/9/e/b/9ebce375a5d242af0392a6ef0cc8dbd7f9651e7d.png)
Thus, we have![\[\frac{5n}{16} > v_5(P(n)) \geq v_7(P(n)) > \frac{8n}{21} - \log_7(n) - 1\]](http://latex.artofproblemsolving.com/9/8/c/98c14c3ecb4923096e3762ddc189be32951406fc.png)
![\[\log_7(n) + 1 > \frac{23n}{16 \cdot 21}\]](http://latex.artofproblemsolving.com/4/4/5/445c3cf3bacf1f3481d755ac3632918b4b222c27.png)
![\[7n > 7^{\frac{23n}{16 \cdot 21}}\]](http://latex.artofproblemsolving.com/a/7/4/a745c9dfc41a112cff0b333d80749ac4713b0fa8.png)
Since the RHS grows faster than the LHS, 
doesn't satisfy the above, so we must have 
.
and 
can be manually calculated, giving 
as the possible candidates, of which only 
work, so we are done.
,
.
.![\[k = \sum_{i=1}^{\infty} \frac{k}{2^i} \geq \sum_{i=1}^{\infty} \left\lfloor\frac{k}{2^i}\right\rfloor = v_2(k!) = \frac{n(n-1)}{2}\]](http://latex.artofproblemsolving.com/d/b/f/dbf1dd3f23db32cb7478c44379ce5510e252032d.png)
Hence we obtain![\[k! \geq \frac{n(n-1)}{2}! \Longrightarrow (2^n-1)(2^n-2)(2^n-4)\cdots(2^n-2^{n-1}) \geq \frac{n(n-1)}{2}!\]](http://latex.artofproblemsolving.com/6/c/a/6cab87f1f6870738beed60c17595c98db1fc87e8.png)
However it is obvious that![\[2^{n^2-1} \geq (2^n-1)(2^n-2)(2^n-4) \cdots (2^n-2^{n-1})\]](http://latex.artofproblemsolving.com/9/f/d/9fde65b1103d659ec97e84c1e80d7e77c7203ae7.png)
the 
appears since 
,![\[2^{n-1} \cdot 4^{\frac{n(n-1)}{2}} = 2^{n^2-1} \geq \frac{n(n-1)}{2}!\]](http://latex.artofproblemsolving.com/7/b/0/7b0a9bbf5b8a9e24802c3519f57f60c9f3e00622.png)
we must have 
,
.
it suffices to prove![\[2^{17} = 2^5 \cdot 4^{15-9} < 10(11)(12)(13)(14)(15)\]](http://latex.artofproblemsolving.com/b/d/6/bd65193dccdda2619af03f8b7c9f1cfe8750917d.png)
However this is obvious since 
and 
.
to obtain the values above (note 
is easy since 
implying 
which is absurd).
:
- this is a corollary of the fact that 
:
.
A cool thing we can do from here is take factorials of both sides (since we already have an expression for 
)
From here, we can rather quickly verify that 
for all 
(induction suffices), therefore all that remains is to check 
.
Meanwhile by LTE and Legendre's 
Thus, 
Testing yields the only solutions 
