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and all positive integers 
and 
satisfying 
is chosen. Prove that there exists a prime number that divides infinitely many terms of the sequence 
, where![\[
b_k = a^{k^k} \cdot 2^{2^k - k} + 1.
\]](http://latex.artofproblemsolving.com/1/7/5/1751a60482d729a36c71b77ac9c978e724f40da0.png)
is interesting if 2018 divides 
(the number of positive divisors of ). Determine all positive integers 
such that there exists an infinite arithmetic progression with common difference whose terms are all interesting.
and 
.
print("this does not import turtle")
then 
for every integers positive 
, and if 
is odd and 
satisfying 
then 
.
and 
.
.
dividing 
equals 
since ![$ 11\|(1990^{1991^{2}}) - ( - 1992) = 1991 [(\sum_{i = 0}^{1991^{2} - 1}( - 1990)^{i}) + 1].$](http://latex.artofproblemsolving.com/6/b/4/6b46080400e56f36e1761a4e716379458835b0b2.png)
and we are done 
then 
is a prime. By Lifting The Exponent lemma ![\[\begin{aligned} v_a \left( (a-1)^{a^{a+1}}+ (a+1)^{a^{a-1}} \right) & = v_a \left( (a-1)^{a^2})^{a^{a-1}}+ (a+1)^{a^{a-1}} \right) \\ & = v_a \left( (a-1)^{a^2}+a+1 \right)+ v_a(a^{a-1}) \\ & = a-1+v_a \left( (a-1)^{a^2}+a+1 \right) \end{aligned} \]](http://latex.artofproblemsolving.com/c/6/1/c6158650ac6f528127fb22b5f676cdbb2bb89b3b.png)
so 
.
. We obtain 
.
of for which 
divides the number ![\[ 1990^{1991^{1992}} + 1992^{1991^{1990}}.\]](http://latex.artofproblemsolving.com/c/0/6/c06d8be837b5996a8c7600460962ce0a74e3b471.png)
. Hence by LTE, 
It remains to find 
. We re-write this as 
. The second one is 
, so we check 
. It follows that 
. Thus, 
and 
and then expanding using the binomial theorem. The "ones" cancel out and we are left with giant powers of 
. Observing the second to last term tells us that the minimal degree of a is indeed .
Since may be prime factorized as 
,
Since 
and 
, the Lifting the Exponent Lemma yields
Note that 
Note that 
, thus 
and , the Lifting the Exponent Lemma yields
Note that 
Note that 
, thus 
, as desired. 
. We can rewrite the expression as ![$$\nu_{1991}(1990^{1991^{1992}} + 1992^{1991^{1990}}) = \nu_{1991}([1990^{1991^2}]^{1991^{1990}} + 1992^{1991^{1990}}$$](http://latex.artofproblemsolving.com/2/4/d/24d4211b182a0b45a8b42cd460fb0d0151802e87.png)
![$$= \nu_{1991}([1990^{1991^2}]^{1991^{1990}} - (-1992^{1991^{1990}})).$$](http://latex.artofproblemsolving.com/3/4/0/340954137ef8866295548bb4deabe202c9013e53.png)
By the LTE lemma, this is equal to 
The finishing step is to realize that 
which implies that the second 
must be equal to exactly 1, hence the result.
.
and 
and thus we are ready to use LTE
both side,we get
Similarly,
which implies 
of for which divides the number 
.Alternately we will prove that 
.We will be using Fermats little theorem and Eulers theorem repeatedly,so we will not state the time of their usages.
Again by standard modulo work we get 
divides the expression 
and thus 
of the expression is .
Similarly we get that
Since 
we can simply ignore the 
condition and write 
Note that 
is a multiple of 11 since the first term is 1 mod 11 and the second is -1 mod 11, so by LTE we have 
We can also compute that 
so 
and the of the entire thing is .
We do some more tedious calculation to see that 
, so this is also 1991.
, we don't actually need to check it for 
since it will also be 
as 
. So we automatically get 
. Many of the above posts actually calculated it.
(
), 
by binomial expansion.
, as we know it will be ..
. We can use LOE to see that 
and 
. For 
. Since 
if 
and 
and and 
are prime, we can see our answer of .
and use LTE on both of those numbers.
. We can use LTE to see that and . For . Since if and and and are prime, we can see our answer of . 
.
is odd and 
but 
and 
so we can apply LTE. It follows that 
so 
. Note we can follow a similar process to get that 
so the answer is .
and 
. We want to find![\[\nu_{1991}(A^{1991^{1990}}+B^{1991^{1990}})\]](http://latex.artofproblemsolving.com/a/0/c/a0c34ae8670d095ffd29cba026a936c6b66f2e65.png)
, we have![\[\nu_p(A^{1991^{1990}}+B^{1991^{1990}})=\nu_p(A+B)+\nu_p(1991^{1990}) = 1991\]](http://latex.artofproblemsolving.com/c/9/c/c9c3ea63162049f8a8c985bba644e8458c143dd2.png)
.
.
.
,
.
.
.
. 
then for every integers positive , and if is odd and satisfying then . and ..
into 
. Then note that 
. Then by LTE 
. We can check easily that 
.
. Since 
, we have 
. Thus, 
. We can easily check that 
is at least similarly, so .
denote the integer in question; we have ![\[N = \left(1990^{1991^2}\right)^{1991^{1990}} + 1992^{1991^{1990}}.\]](http://latex.artofproblemsolving.com/6/6/f/66f535c00054918d7dcdae5da944335f3edc12c5.png)
Noting that 
, by LTE we obtain ![\[\nu_{11}(N) = \nu_{11}\left(1990^{1991^2} + 1992\right) + \nu_{11}(1991^{1990}) = 1 + 1990 = 1991,\]](http://latex.artofproblemsolving.com/0/1/9/019c5ad4e6790e8e757deea98cc7095c5d5ae405.png)
since ![\[\nu_{11}\left(1990^{1991^2} + 1^{1991^2}\right) = \nu_{11}(1991) + \nu_{11}(1991^2) = 3\]](http://latex.artofproblemsolving.com/f/b/a/fbab5b664e3666181aa175b3ed43476ea6054e8e.png)
and ![\[\nu_{11}\left(1990^{1991^2} + 1992\right) = \nu_{11}\left(\left(1990^{1991^2} + 1\right) + 1991\right) = \min(3, 1) = 1.\]](http://latex.artofproblemsolving.com/8/0/f/80fa631fcd71f9a048fdd01d02c1287fe3b179fb.png)
Similarly, 
for the exact same reasons. Thus the largest such integer is . 
denotes the highest power of which divides . Using 
,
, similarly 
.. OG!
and integer 
the following holds.
.
Notice 
and 
Hence, by LTE, 
Notice 
so 
and 
Similarly for 
we can conclude that 
divides our number, so 
.
Note that (using 
)
and similarly 
,
,
.
where 
because 
,
,
and 
.
,![\[1991^{1991^2}-\dots+\binom{1991^2}{1}\cdot 1991-1+1991+1\equiv 1991\cdot 2\pmod{1991^2}.\]](http://latex.artofproblemsolving.com/e/6/f/e6f200b67b69cea12e9f35e48761eabb5b3a4b09.png)
.
Similarly, 
,
![\[ v_{11}(1990^{1991^{1992}} + 1992^{1991^{1990}} = v_{11}((1990^{1991^2})^{1991^{1990}} + 1992^{1991^{1990}}) = v_{11}(1990^{1991^2} + 1992) + 1990 \]](http://latex.artofproblemsolving.com/0/b/2/0b2cc4977ca1407d869e77ce247a8bdde16fe45d.png)
Note that 
.![\[ v_{11}(1992^{1991^2} - 1992) = v_{11}(1992^{1990 \cdot 1992} - 1) = v_{11}(1991) + v_{11}(1990 \cdot 1992) = 1 \]](http://latex.artofproblemsolving.com/5/1/3/513d7b20f7f6244786fa258d75840352c0ac7c6e.png)
Thus, 
,
.![\[v_p \left((1990^{1991^2})^{1991^{1990}} + (1992)^{1991^{1990}}\right) = v_p(1990^{1991^2}+1992) + 1990.\]](http://latex.artofproblemsolving.com/c/1/c/c1c58eec150d55ea5c95bbcd943de488695c2b30.png)
,
,
divides 
,
.
and 
Now, since the prime factorization of 
it suffices to find the smaller of 
and 
It is easy to verify that 
so the Lifting the Exponent Lemma gives us
Analogously, we have 
Note that
so we have 
But it is not difficult to deduce that 
so it follows that 
Therefore, the answer is 