# 10 Math Problems That Look Easy But Immensely Difficult to Solve

There are thousands of math problems that are difficult for a common person to understand even though mathematicians may find them easy to solve.  On the other hand, there are math problems that look really easy that even a middle student would understand the way they are stated, but their solution or proof is immensely difficult. Yes, such math problems exist and below are the some of the most well known.

1. Squaring the Circle
Squaring the circle is one of the classic math problems proposed by Geometers.
It was a challenge to use compass and straightedge to construct  a square with the same area as a given circle in a finite number of steps. Although the circle to square approximation was known since the time of the ancient Babylonian mathematicians, it was Anaxagoras (c. 510 – 428 BC) who was the first to be recorded in history to work on the problem.

In 1882, Ferdinand von Lindemann proved that $\pi$ was transcendental. The consequence of this is the impossibility of squaring the circle.

2. Twin Prime Conjecture

Prime numbers are positive integers that are divisible only by $1$ and itself. The first 10 prime numbers are $2, 3, 5, 7, 11, 13, 17, 19, 23$, and $29$. Twin Primes are pair of primes with difference $2$. The pairs $(3,5)$, $(5,7)$, $(11,13)$ and $(17,19)$ are twin primes. The greatest twin primes as of this writing are

$3756501695685 \times 2^{66669} - 1$

and

$3756501695685 \times 2^{66669} + 1$

both primes consisting $200,700$ digits (reference). Now the question is how many twin primes are there?

Although the infinitude of primes was proved as early as the time of Euclid, the Twin Prime Conjecture is still one of the most elusive math problems until now. Although mathematicians believe that the conjecture is true, the answer still remains a mystery.

3. Goldbach’s Conjecture (2 conjectures)

This math problem was proposed by German mathematician Christian Goldbach to Leonhard Euler in 1742 and is still unsolved until now. It states that every even integer greater than $2$ can be expressed as a sum of two primes. The following are examples: $4 = 2 + 2$, $6 = 3 + 3$, $8 = 3 + 5$,  and $10 = 5 + 5 = 3 + 7$. The image below shows that this is true up to $28$.

Click image to enlarge (Image Credit: Wikipedia)

In fact, as of this writing, it has been tested to be true up to $4 \times 10^{18}$ but is still unproven in general.

Another version of this math problem also exists, the Goldbach’s weak conjecture. It states that every positive integer greater than $5$ can be expressed as sum of three primes given that a prime may be used more than once in the same sum

4. Fermat’s Last Theorem

Everybody who has taken high school mathematics is familiar with the Pythagorean Theorem. This theorem has hundreds of proofs. The Pythagorean Triples $(a, b, c)$ where $a, b, c$ are lengths of right triangles and satisfy the Pythagorean Theorem equation

$a^2 + b^2 = c^2$.

Although we can find infinitely many triples (e.g. mulitply $( 3, 4, 5)$ by any positive integer) that satisfy the Pythagorean Theorem, the Fermat’s Last Theorem states that there are no positive integers $(a, b, c)$. that satisfy the equation with higher degree than $2$. That is, for every triple $(a, b, c)$, the following inequalities are always true.

$a^3 + b^3 \neq c^3$

$a^4 + b^4 \neq c^4$

$a^5 + b^5 \neq c^5$

$\cdots$

$a^n + b^n \neq c^n$.

In general, it says that there is are no positive integers $a$, $b$, $c$ that satisfy the equation $a^n + b^n = c^n$, for any integer $n$ greater than $2$.

This math problem was proposed by Pierre de Fermat in 1637, and after more than 300 years and the many incorrect proofs published, Andrew Wiles finally solved the enigma in 1995. It took him 8 years to solve the problem!

5. Catalan’s Conjecture

The Catalan’s conjecture states that only $2^3$ and $3^2$ are the powers of two natural numbers whose values 8 and 9 are consecutive. This was conjectured by an Eugène Charles Catalan in 1844 and proven in 2002 by Preda Mihăilescu.This means that aside from the pair above, there is no solution to the equation

$x^a - y^b = 1$

where $x, a, y, b >$ 1.

6. Four Color Theorem

The Four Color Theorem states that given a “map” with contiguous regions, the minimum number of colors that you can place where no adjacent of the same color is 4.

It was proposed by Francis Guthrie in 1852 and was proved by Kenneth Appel and Wolfgang Haken in 1976. It was the first major theorem to be proved using a computer.

7. Collatz Conjecture

Take any natural number. If even, divided by $2$. If odd, multiply by $3$ and add $1$. The Collatz conjecture states that no matter what number you choose at first, doing this repeatedly will eventually result to 1.  The conjectures is still unsolved to this day.

8. Perfect Number Conjectures (2 conjectures)

A perfect number is a positive integer that is equal to the sum of its proper divisors (the divisors less than it). The smallest perfect numbers is $6$: its divisors are $1$, $2$, and $3$ and $1 + 2 + 3 = 6$. The next perfect number is $28$ ($1 + 2 + 4 + 7 + 14 = 28$. Perfect numbers are very rare. As of this writing, there are only $48$ known perfect numbers, and no one knows if there are infinitely many of them. All known perfect numbers are also even and no one also knows if an odd perfect number exists.

The problems above show one aspect of mathematical beauty. That sometimes, the simple math problems (to state) are the most difficult to solve.

## 5 thoughts on “10 Math Problems That Look Easy But Immensely Difficult to Solve”

1. I can square a circle. 🙂 j/k

This is a nice collection of very difficult math problems, some already solved and some not.

I was confused by the Collatz Conjecture, as I didn’t understand how an odd number multiplied by three, plus one, could add up to one in the end. What you may consider to clarify is that you are looking at the SERIES of numbers built up from a previous n, not just any singular n value. So, for n = 3, you start with the odd number 3, which gives you 3*3+1 = 10. 10 is now even, so that is 10/2 = 5. 5 is odd, which gives 5*3+1, which is 16, and so on.

• See if you choose 6 , then since it is even , you get 3 and now its odd so you get 10 , then you get 5 and then you get 16 and then you get 8 and then you will get 4 and then you will get 2 and lastly you will reach 1. So its completely true…
sequence will goes like 6, 3, 10, 5, 16, 8, 4, 2, 1.

2. Good list, but I think the title is a bit misleading. These problems do NOT look easy, at least not to me. I don’t think I’ve ever found a problem that was difficult for me to solve that didn’t look difficult at first, so maybe it’s just me. These problems are easy to understand though, if that’s what you mean.