\(\begin{bmatrix} a && b \\ c && d \end{bmatrix}*\begin{bmatrix} a && b \\ c && d \end{bmatrix}=\begin{bmatrix} a^2+b*c &&a*b+b*d \\ c*a+d*c && d^2+c*b \end{bmatrix}= \begin{bmatrix} a^2+bc && b*(a+d) \\ c*(a+d) && d^2+bc \end{bmatrix}=\begin{bmatrix} 0 && 0 \\ 0 && 0 \end{bmatrix}\)

(The multiplication of the matrices is derived from the formula: \((A*B)i,j =\sum_{k=1}^{n}a(i,k)*b(k,j)\)

so we have a set of equations:

1. a²+bc=0

2. b*(a+d)=0

3. c*(a+d)=0

4. d²+bc=0

using the first and the fourth equations, we can find that :(a²+bc)-(d²+bc)=(0)-(0)=0=a²-d²=(a-d)*(a+d). therefore, we can divide our answer to two cases:

first case- a=d

that means b*(a+d)=0=b*(2a) and c*(a+d)=0=c*(2a). now we have a new set of equations:

1. a²+bc=0

2. b*(2a)=0

3. c*(2a)=0

we can make it simpler:

1. a²+bc=0

2. b*a=0

3. c*a=0

now we can divide that case to two different cases:

a=0:

now we have 1 equation:

1. a²+bc=0=0²+bc=bc=0. that means b or c must be equal to 0 (that or is a mathematical or: it means if both of them are 0 it works too)

that means the matrices that maintain the equations are of the following form:

\(\begin{bmatrix} 0 && b \\ c && 0 \end{bmatrix}\)(where c*b=0)

a is not 0:

that means we can divide the second and the third equations by a and get the set of equations:
 

1. a²+bc=0

2. b=0

3. c=0

that means a²+bc=a²+0*0=a². but that means a=0, and that doesnt make any sense because we know that a CANNOT be 0. So there is no solution

second case- d=-a

now we can derive a simpler set of equations:

1. a²+bc=0

2. b*(a-a)=0=b*0=0

3. c*(a-a)=0=c*0=0

now we have one equation- a²+bc=0:

a²+bc=0/ -a²= 

bc=-a²

that gives us the next set of matrices: \(\begin{bmatrix} (-1)^n\sqrt{-bc} && b \\ c && -(-1)^n\sqrt{-bc} \end{bmatrix} \)

(where n is a natural number, couldnt find the plus minus sign). but as you can see, every matrice from the first set of matrices is also a member of the second set of matrices, therefore we can simply say that every matrice that is a member of the second set is a solution (keep in mind that every matrice from that set is a solution)

but we dont want A=0 as a solution, so we simply have to add that b*c is not 0.

solution: \(\begin{bmatrix} (-1)^n\sqrt{-bc} && b \\ c && -(-1)^n\sqrt{-bc} \end{bmatrix} \)

(where b*c is not 0)

I'll prove that for every natural , n>1, for given positive integers a₁, a₂, .....,a_n and S={x₁a₁+x₂a₂+......+x_na_n: a₁, a₂, ....,a_n are positive integers} S contains infinitely many primes if and only if no positive integer divides all n integers (a₁, a₂,......,a_n)

proof:

suppose there is a positive integer P, so that a_i is divisible by P, for every 0 1a ₁+x ₂a ₂+......+x _na _n is always divisible by P, and because there arent infinitely many primes that are divisible by any positive integer P (unless P=1, but we know P isnt 1) there arent infinitely many primes in S.

I'll prove that for every natural , n>1, for given positive integers a₁, a₂, .....,a_n and S={x₁a₁+x₂a₂+......+x_na_n: a₁, a₂, ....,a_n are positive integers}, there exists a positive integer D, so that for every D 1, x ₂, x ₃,......,x _n so that  x ₁a ₁+x ₂a ₂+......+x _na _n=y if no positive integer divides all n integers (a ₁, a ₂,......,a _n).

proof:

The proof is by induction- Suppose the positive integer P divides all positive integers a₁, a₂, .....,a_n-1, and that after dividing each of the integers by P no other integers divides them all. Lets define b_i=a_i/P for every 0 1 1, x ₂, x ₃,......,x _n-1 so that x ₁b ₁+x ₂b ₂+......+x _n-1b _n-1=y Therefore, for every y that is divisible by P and bigger than P*D ₁ there exists positive integers x ₁, x ₂, x ₃,......,x _n-1 so that  x ₁a ₁+x ₂a ₂+......+x _n-1a _n-1=y. Suppose a _n and P are coprime. Therefore, we can express every number of the form x ₁a ₁+x ₂a ₂+......+x _na _n using the sum r ₁*P+r ₂*a _n. Using the chicken-nugget theorem (Search it up, its real) we can infer that there exists an integer D ₂ so that for every D ₂ 1 and r ₂ so that r ₁*P+r ₂*a _n=y.

(If a_n and P are NOT coprime that means another positive integer, F, divides both of them, and that means F divides all of the integers (a₁, a₂,......,a_n), but that is a contradiction to what we said in the beginning- no positive integer divides all of the integers).

And therefore, if the theorem

for every natural , n>1, for given positive integers a₁, a₂, .....,a_n and S={x₁a₁+x₂a₂+......+x_na_n: a₁, a₂, ....,a_n are positive integers}, there exists a positive integer D, so that for every D 1, x ₂, x ₃,......,x _n so that  x ₁a ₁+x ₂a ₂+......+x _na _n=y if no positive integer divides all n integers (a ₁, a ₂,......,a _n).

Is true for n-1, it is also true for n.

I'll complete the theorem by proving it for n=2: suppose a₁ and a₂ are positive integers. if no positive integers divides both of them, then they are coprime integers, and using the chicken nugget theorem we can infer there exists a positive integer D₁ so that for every D₁ 1a ₁+x ₂a ₂=y for some positive integers x ₁ and x ₂.

Therefore, the theorem is true for n>1.

We know there are infinitely many primes, and we know that for every positive integer D there exists a FINITE number of primes between 1 and D, So there are infinitely many primes that are equal or bigger than D.  from the following theorem, we can infer Every prime that is bigger than D will be in the set S, and therefore, there are infinitely many primes in S IF AND ONLY IF no positive integer divides a₁, a₂,......,a_n.

Now, this was just a generalization of your question, and I'll use it to answer your question-

Suppose b and c are coprime. Therefore, there exists a positive integer D so that for every D 1b+x ₂c=y for some positive integers x ₁ and x ₂. therefore, for every a+D 1b+x ₂c=y for some positive integers x ₁ and x ₂. That means there are infinitely many primes in S.

so if b and c are coprime, we can infer that there are infinitely many primes in S. We also know that if there exists a positive integer P so that it divides a, b and c There is a finite number of primes in S.

The question is: what if there exists a positive integer P so that P divides b, and c, but not a? that is equivalent to the question: Is there an arithmetic sequence a_n=a+bn where a and b are coprime and that contains a finite number of primes?

unfortunately i cant answer that question, BUT i have this cool generalization (?) of the question, and i love generalizations, so i'll keep it although its useless.

EDIT: The answer to the question "Is there an arithmetic sequence a_n=a+bn where a and b are coprime and that contains a finite number of primes?" is NO, and therefore S contains infinitely many primes only and only if no positive integer divides both a, b, and c. But We have to prove it using Dirichlet's theorem on arithmetic progressions, So i cant really prove it.

Who gave you this question? because whoever did must be a horrible person.

Yes, it is. You can multiply any 2 numbers by eachother. Doesnt matter if they're rational, or irrational, or complex, or real. doesnt matter. You can multiply them.

You truly are the LaTeX master

Although i believe the expression cant be

(x-(x²+4)^1/2)/2 because that means its negative, and im quite sure it cant be negative. I cant prove it properly right now, but i believe there is only one answer to the question.

Ok. I think i got this-

Lets call the expression f(x). we know that x+1/f(x)=f(x). We can multiply by f(x) and get-

x*f(x)+1=f(x)²     |  subtract x*f(x)

f(x)²-x*f(x)=1    | add x²/4 

(f(x)-x/2)²=1+x²/4

f(x)=x/2+(1+x²/4)^1/2.

Im terrible at finding integrals, so i used an integral calculator:

its integral is:

1/4 (x (sqrt(x^2 + 4) + x) + 4 sinh^(-1)(x/2)) + constant

The answer is 2

Whats the point of your first question? couldnt you just call the monstrosity on the right side of the equation 'x'?

and I didnt really get your second question.

\(\lim_{n\rightarrow infinity}{{(\sqrt[n]{|{(-1)}^{n}*(1-(\frac{15}{n}))|})}^{n}}^{2}= \lim_{n\rightarrow infinity}{|{(-1)}^{n}*(1-(\frac{15}{n}))|}^{2}=(\lim_{n\rightarrow infinity}{|{(-1)}^{n}*(1-(\frac{15}{n}))|})^2=1^2=1.\)

But the function is 4/e*e^x-x-1 isnt it? isnt it f(2)=4e-3?

There is also a generalization of this question:

aⁿ+bⁿ/(ab)^n-1+1 is an integer k. Prove that k=cⁿ where c is also an integer (a, b and n are all integers)

"your definately not 4."

what?

I dont think finding f(2) is possible....

f'(x)=x+f(x)--------->

f'''(x)=(x)''+f''(x)=f''(x)----->

f''(x)=c₁*e^x------->

f(x)=c₁*e^x+c₂*x+c₃

f(1)=c₁*e+c₂+c₃=2-------------->

c₃=2-c₂-c₁*e

f(2)=c₁*e²+c₂*2+c₃=

c₁*e²+c₂+2-c₁*e

edit: OH SHOOT I FORGOT SOMETHING

f'(x)=c₂+c₁*e^x=x+c₃+c₂*x+c₁*e^x  | subtract c₁*e^x ----->

c₂=x+c₃+c₂*x. therefore, c₂=-1(the only option. that means c₃=c₂=-1 and that means

f(1)=c₁*e-2=2 therefore c₁=4/e

that means f(2)=4*e-2-1=4e-3.

I didnt understand the guest's way, so ill solve it my way:

P(k)=the number of times the digit k appears between 0 and 999,999. P(0)*0+P(1)*1+P(2)*2+....+P(9)*9+1= the sum we need to find (i replaced 1,000,000 with 0, all we need to do is add 1 to the final solution.

all the numbers between 0 and 999,999 are also every possible combination of 6 digits (instead of 1 we write 0000001, instead of 2 we write 000002. Yes, a large number of zeroes is added, but that doesnt change the solution because x+n*0=x+0=x)

Here's a link to a similar question someone asked about the number of times a digit appears between 1 and 10ⁿ(n is a positive integer):

https://web2.0calc.com/questions/how-many-3s#r21

. i told him that between 0 and 10ⁿ-1 ,every digit (if we write it the way i write the numbers) appears n*10^n-1 times, meaning every digit except for 0 appears n*10^n-1 (but we dont care about the zeroes). what we want to find is the number of times every digit (except for 0) appears between 0 and 999,999=10⁶-1. so, the answer is 6*10^6-1=6*10⁵.

so, P(k)=6*10⁵ for every k that is between 1 and 9.

P(0)*0+P(1)*1+P(2)*2+....+P(9)*9+1=P(1)*1+P(2)*2+....+P(9)*9+1=6*10⁵(1+2+3+4....+9)+1=6*10⁵*45+1=27,000,000+1=

27,000,001

I think i got the answer.

The sequence can be described by the following formula: a_n+2=a_n*(n+2), a₁=1 and a₂=2

That means the next 3 terms are: 384*10=3840, 945*11=10395, 3840*12=46080

But there is something else i want to add about this type of questions:

I feel its a little bit silly, because you could always find another pattern or another way. Of course, if you need to answer that type of questions in a test, you cant write that, you'll have to do what they expect you to do. But i feel its wrong NOT to mention what im going to.

I once heard a joke about a mathematician that has an interesting way of answering those questions: he simply writes "19" when he's asked to find the next term. He justifies that with one of the theorems of a famous mathematician called Lagrange. And in my opinion he's right. Saying that there is a finite amount of answers is silly, because there isnt. There is always a way to make order out of what seems to be chaos.

I dont know if you love math or care about it, but if you do, remember it.