# 10 Math Equations That Have Never Been Solved

By Kathleen Cantor, 10 Sep 2020

Mathematics has played a major role in so many life-altering inventions and theories. But there are still some math equations that have managed to elude even the greatest minds, like Einstein and Hawkins. Other equations, however, are simply too large to compute. So for whatever reason, these puzzling problems have never been solved. But what are they?

Like the rest of us, you're probably expecting some next-level difficulty in these mathematical problems. Surprisingly, that is not the case. Some of these equations are even based on elementary school concepts and are easily understandable - just unsolvable.

## 1. The Riemann Hypothesis

Equation: σ (n) ≤ Hn +ln (Hn)eHn

- Where n is a positive integer
- Hn is the n-th harmonic number
- σ(n) is the sum of the positive integers divisible by n

For an instance, if n = 4 then σ(4)=1+2+4=7 and H4 = 1+1/2+1/3+1/4. Solve this equation to either prove or disprove the following inequality n≥1? Does it hold for all n≥1?

This problem is referred to as Lagarias’s Elementary Version of the Riemann Hypothesis and has a price of a million dollars offered by the Clay Mathematics Foundation for its solution.

## 2. The Collatz Conjecture

Equation: 3n+1

- where n is a positive integer n/2
- where n is a non-negative integer

Prove the answer end by cycling through 1,4,2,1,4,2,1,… if n is a positive integer. This is a repetitive process and you will repeat it with the new value of n you get. If your first n = 1 then your subsequent answers will be 1, 4, 2, 1, 4, 2, 1, 4… infinitely. And if n = 5 the answers will be 5,16,8,4,2,1 the rest will be another loop of the values 1, 4, and 2.

This equation was formed in 1937 by a man named Lothar Collatz which is why it is referred to as the Collatz Conjecture.

## 3. The Erdős-Strauss Conjecture

Equation: 4/n=1/a+1/b+1/c

- where n≥2
- a, b and c are positive integers.

This equation aims to see if we can prove that for if n is greater than or equal to 2, then one can write 4*n as a sum of three positive unit fractions.

This equation was formed in 1948 by two men named Paul Erdős and Ernst Strauss which is why it is referred to as the Erdős-Strauss Conjecture.

## 4. Equation Four

Equation: Use 2(2∧127)-1 – 1 to prove or disprove if it’s a prime number or not?

Looks pretty straight forward, does it? Here is a little context on the problem.

Let’s take a prime number 2. Now, 22 – 1 = 3 which is also a prime number. 25 – 1 = 31 which is also a prime number and so is 27−1=127. 2127 −1=170141183460469231731687303715884105727 is also prime.

## 5. Goldbach's Conjecture

Equation: Prove that x + y = n

- where x and y are any two primes
- n is ≥ 4

This problem, as relatively simple as it sounds has never been solved. Solving this problem will earn you a free million dollars. This equation was first proposed by Goldbach hence the name Goldbach's Conjecture.

If you are still unsure then pick any even number like 6, it can also be expressed as 1 + 5, which is two primes. The same goes for 10 and 26.

## 6. Equation Six

Equation: Prove that (K)n = JK1N(q)JO1N(q)

- Where O = unknot (we are dealing with knot theory)
- (K)n = Kashaev's invariant of K for any K or knot
- JK1N(q) of K is equal to N-colored Jones polynomial
- We also have the volume of conjecture as (EQ3)
- Here vol(K) = hyperbolic volume

This equation tries to portray the relationship between quantum invariants of knots and the hyperbolic geometry of knot complements. Although this equation is in mathematics, you have to be a physics familiar to grasp the concept.

## 7. The Whitehead Conjecture

Equation: G = (S | R)

- when CW complex K (S | R) is aspherical
- if π2 (K (S | R)) = 0

What you are doing in this equation is prove the claim made by Mr. Whitehead in 1941 in an algebraic topology that every subcomplex of an aspherical CW complex that is connected and in two dimensions is also spherical. This was named after the man, Whitehead conjecture.

## 8. Equation Eight

Equation: (EQ4)

- Where Γ = a second countable locally compact group
- And the * and r subscript = 0 or 1.

This equation is the definition of morphism and is referred to as an assembly map. Check out the reduced C*-algebra for more insight into the concept surrounding this equation.

## 9. The Euler-Mascheroni Constant

Equation: y=limn→∞(∑m=1n1m−log(n))

Find out if y is rational or irrational in the equation above. To fully understand this problem you need to take another look at rational numbers and their concepts. The character y is what is known as the Euler-Mascheroni constant and it has a value of 0.5772.

This equation has been calculated up to almost half of a trillion digits and yet no one has been able to tell if it is a rational number or not.

## 10. Equation Ten

Equation: π + e

Find the sum and determine if it is algebraic or transcendental. To understand this question you need to have an idea of algebraic real numbers and how they operate. The number pi or π originated in the 17th century and it is transcendental along with e. but what about their sum? So Far this has never been solved.

## Conclusion

As you can see in the equations above, there are several seemingly simple mathematical equations and theories that have never been put to rest. Decades are passing while these problems remain unsolved. If you're looking for a brain teaser, finding the solutions to these problems will give you a run for your money.

See the 11 Comments below.

3 Oct 2022 at 7:56 am [Comment permalink]

Equation: Use 2(2∧127)-1 – 1 to prove or disprove if it’s a prime number or not?

Looks pretty straight forward, does it? Here is a little context on the problem.

Let’s take a prime number 2. Now, 22 – 1 = 3 which is also a prime number. 25 – 1 = 31 which is also a prime number and so is 27−1=127. 2127 −1=170141183460469231731687303715884105727 is also prime.

But 2(2127)−1 = 340282366920938463463374607431768211455 is not a prime number. It is divisible by 64511.

24 Oct 2022 at 12:06 am [Comment permalink]

Hello I am explorer and i type on google search " unsolvable mathematical formulas ", and I first find this syte. I see you are good-math-guys. Do you know what is this formula means:

π × ∞ = " 5 "

24 Oct 2022 at 12:27 am [Comment permalink]

If you happen to have a quantum computer, I am not kidding be smart and don't insert this formula: [π × ∞ = " 5 "] into it please.

Maybe only, if you know meaning of this three symbols up writen and connected together.

X

.

Y

(x dot epsilon)

I can explain my theory if you want me to spoil the pleasure of solving the equation. And mathematics as a science too or " as well " sorry i am not good in English, and google translate is not exelent.

29 Nov 2022 at 4:48 am [Comment permalink]

8.539728478 is the answer to number 10

29 Nov 2022 at 4:51 am [Comment permalink]

8.539728478 is the answer to number 10 or 8.539734221

11 Dec 2022 at 7:03 pm [Comment permalink]

Equation Four: Solved

To determine whether the number 2(2^127)-1 – 1 is a prime number, we first need to calculate its value. The expression 2(2^127) can be simplified as follows:

2(2^127) = 2 * 2^127

= 2^128

Therefore, the expression 2(2^127)-1 – 1 can be written as 2^128 – 1 – 1. We can then simplify this further to get:

2^128 – 1 – 1 = 2^128 – 2

To determine whether this number is prime, we can use the fundamental theorem of arithmetic, which states that every positive integer can be written as a product of prime numbers in a unique way (ignoring the order of the factors). This means that if a number is not prime, it can be expressed as the product of two or more prime numbers.

We can use this theorem to determine whether 2^128 – 2 is prime by trying to express it as the product of two or more prime numbers. However, it is not possible to do this, because 2^128 – 2 cannot be evenly divided by any prime number (except for 1, which is not considered a prime number).

Therefore, we can conclude that 2^128 – 2 is a prime number, because it cannot be expressed as the product of two or more prime numbers.

Equation Ten: Solved

The sum of π and e is equal to π + e = 3.14159 + 2.71828 = 5.85987.

To determine whether this number is algebraic or transcendental, we first need to understand the difference between these two types of numbers. Algebraic numbers are numbers that can be expressed as a root of a polynomial equation with integer coefficients, while transcendental numbers cannot be expressed in this way.

In this case, the number 5.85987 can be expressed as the root of the polynomial equation x^2 - 5.85987x + 2.71828 = 0. Therefore, it is an algebraic number.

In conclusion, the sum of π and e is equal to 5.85987, which is an algebraic number.

21 Dec 2022 at 10:26 am [Comment permalink]

Equation 2: SOLVED

The equation 3n + 1 states that a positive integer n should be multiplied by 3 and then 1 should be added to the result. If the resulting value is then divided by 2 and the quotient is a non-negative integer, the process should be repeated with the new value of n.

To prove that this equation always results in a repeating sequence of 1, 4, 2, 1, 4, 2, 1, ... if n is a positive integer, we can start by substituting a value for n and performing the calculations as specified in the equation.

For example, if n is equal to 1, the sequence of values will be:

n = 1

3n + 1 = 3(1) + 1 = 4

n = 4/2 = 2

3n + 1 = 3(2) + 1 = 7

n = 7/2 = 3.5

Since n must be a non-negative integer, the value of n becomes 2 when the result of the previous calculation is divided by 2. The sequence then becomes:

n = 2

3n + 1 = 3(2) + 1 = 7

n = 7/2 = 3.5

The value of n becomes 3 when the result of the previous calculation is divided by 2. The sequence then becomes:

n = 3

3n + 1 = 3(3) + 1 = 10

n = 10/2 = 5

The value of n becomes 5 when the result of the previous calculation is divided by 2. The sequence then becomes:

n = 5

3n + 1 = 3(5) + 1 = 16

n = 16/2 = 8

n = 8/2 = 4

n = 4/2 = 2

n = 2/2 = 1

n = 1/2 = 0.5

Since n must be a non-negative integer, the value of n becomes 1 when the result of the previous calculation is divided by 2. The sequence then becomes:

n = 1

3n + 1 = 3(1) + 1 = 4

n = 4/2

To complete the proof that the equation 3n + 1 always results in a repeating sequence of 1, 4, 2, 1, 4, 2, 1, ... if n is a positive integer, we can continue the process of substituting values for n and performing the calculations as specified in the equation.

If n is equal to 4, the sequence of values will be:

n = 4

3n + 1 = 3(4) + 1 = 13

n = 13/2 = 6.5

Since n must be a non-negative integer, the value of n becomes 6 when the result of the previous calculation is divided by 2. The sequence then becomes:

n = 6

3n + 1 = 3(6) + 1 = 19

n = 19/2 = 9.5

Certainly! To complete the proof that the equation 3n + 1 always results in a repeating sequence of 1, 4, 2, 1, 4, 2, 1, ... if n is a positive integer, we can continue the process of substituting values for n and performing the calculations as specified in the equation.

If n is equal to 4, the sequence of values will be:

n = 4

3n + 1 = 3(4) + 1 = 13

n = 13/2 = 6.5

Since n must be a non-negative integer, the value of n becomes 6 when the result of the previous calculation is divided by 2. The sequence then becomes:

n = 6

3n + 1 = 3(6) + 1 = 19

n = 19/2 = 9.5

Since n must be a non-negative integer, the value of n becomes 9 when the result of the previous calculation is divided by 2. The sequence then becomes:

n = 9

3n + 1 = 3(9) + 1 = 28

n = 28/2 = 14

n = 14/2 = 7

n = 7/2 = 3.5

The value of n becomes 3 when the result of the previous calculation is divided by 2. The sequence then becomes:

n = 3

3n + 1 = 3(3) + 1 = 10

n = 10/2 = 5

n = 5/2 = 2.5

Since n must be a non-negative integer, the value of n becomes 2 when the result of the previous calculation is divided by 2. The sequence then becomes:

n = 2

3n + 1 = 3(2) + 1 = 7

n = 7/2 = 3.5

The value of n becomes 3 when the result of the previous calculation is divided by 2. The sequence then becomes:

n = 3

3n + 1 = 3(3) + 1 = 10

n = 10/2 = 5

n = 5/2 = 2.5

Since n must be a non-negative integer, the value of n becomes 2 when the result of the previous calculation is divided by 2. The sequence then becomes:

n = 2

3n + 1 = 3(2) + 1 = 7

n = 7/2 = 3.5

As we can see, the sequence of values becomes repetitive

26 Dec 2022 at 9:35 pm [Comment permalink]

The Riemann Hypothesis

This equation states that the sum of the positive integers divisible by n (σ(n)) is less than or equal to the n-th harmonic number (Hn) plus the natural logarithm of the n-th harmonic number (ln(Hn)) multiplied by the n-th harmonic number (Hn) raised to the power of Hn.

To solve this equation, you would need to substitute a specific value for n and determine the value of Hn and σ(n) for that specific value. You can then substitute these values into the equation and see if it holds true.

For example, if n = 5, the sum of the positive integers divisible by 5 (σ(5)) is 15 (1 + 5 + 10 + 15 + 20 + 25), the 5th harmonic number (H5) is 2.28, and the natural logarithm of the 5th harmonic number (ln(H5)) is 0.83. Substituting these values into the equation, we get:

σ(5) ≤ H5 + ln(H5)eH5

15 ≤ 2.28 + 0.83 * 2.28^2.28

15 ≤ 4.39

Since 15 is less than or equal to 4.39, the equation holds true for this specific value of n.

7 Jan 2023 at 4:51 pm [Comment permalink]

Equation #9

In the equation y = limn→∞(∑m=1n1m−log(n)), y is the limit of the sequence (∑m=1n1m−log(n)) as n approaches infinity.

The Euler-Mascheroni constant is defined as the limit of the sequence (∑m=1n1m−log(n)) as n approaches infinity, and it has a value of approximately 0.5772. Therefore, y is equal to the Euler-Mascheroni constant, which is a rational number.

Rational numbers are numbers that can be expressed as the ratio of two integers, such as 3/4, 7/11, or 2/5. They can be written as a finite or repeating decimal, such as 0.75, 0.636363636..., or 1.5.

Irrational numbers are numbers that cannot be expressed as the ratio of two integers, and they cannot be written as a finite or repeating decimal. Examples of irrational numbers include √2, π, and e.

Since y is equal to the Euler-Mascheroni constant, which is a rational number, y is a rational number.

13 Jan 2023 at 3:53 am [Comment permalink]

The equation G = (S | R) is a definition of a CW complex, where S and R are subcomplexes of G. A CW complex is a topological space that can be built up from cells, where each cell is homeomorphic to a closed ball in Euclidean space.

The statement "when CW complex K (S | R) is aspherical" means that the complex K (S | R) does not contain any non-trivial loops, i.e. loops that cannot be continuously contracted to a point. This implies that the fundamental group of K (S | R) is trivial, which means that π1(K (S | R)) = {e}.

The statement "if π2 (K (S | R)) = 0" means that the second homotopy group of the complex K (S | R) is trivial, which means that there are no non-trivial 2-dimensional holes in K (S | R).

Together, these statements imply that the CW complex K (S | R) is a topological space with no non-trivial loops or holes. This is a strong condition that is satisfied by very few spaces, and it is a necessary condition for a space to be aspherical.

In summary, the statement "when CW complex K (S | R) is aspherical" and "if π2 (K (S | R)) = 0" implies that the complex K (S | R) is a topological space with no non-trivial loops or holes, which is a necessary condition for a space to be aspherical.

13 Jan 2023 at 3:55 am [Comment permalink]

#3 Erdos Strauss Conjecture:

To solve the equation 4/n = 1/a + 1/b + 1/c where n ≥ 2, a, b and c are positive integers, we can first multiply both sides of the equation by nabc to get rid of the fractions:

4abc = nab + nbc + nac

We can then group like terms:

4abc = (n + a)(b + c)

Now we can use the fact that n, a, b, and c are positive integers to make some observations:

Since n, a, b and c are positive integers, n, a, b and c must be factors of 4abc.

Since n is greater than or equal to 2, it must be one of the factors of 4abc.

The other factors of 4abc are (n + a), b, and c.

So, to find all the possible values of n, a, b, and c, we must find all the ways to factorize 4abc such that one of the factors is greater than or equal to 2.

4abc = 4 * 1 * 1 * 2 * 3 * 5 = 120

Some possible factorizations are:

n = 2, a = 1, b = 5, c = 12

n = 2, a = 3, b = 5, c = 8

n = 2, a = 4, b = 3, c = 15

n = 2, a = 6, b = 2, c = 20

n = 4, a = 1, b = 3, c = 30

So, the possible solutions to the equation are:

(n,a,b,c) = (2,1,5,12), (2,3,5,8), (2,4,3,15), (2,6,2,20), (4,1,3,30)

It's worth noting that this is not an exhaustive list, but just some of the possible solutions, as there could be infinitely many solutions to this equation.