A simple explanation of a/(b+c) + b/(c+a) + c/(a+b) = 4
2025 May 11
See all posts
A simple explanation of a/(b+c) + b/(c+a) + c/(a+b) = 4
You may have at some point seen this math puzzle:
The puzzle has gained some degree of notoriety on the internet
because it looks like it must be either simple, impossible, or
a trick question (solve P = NP? Hahaha, the answer is N = 1, got you!),
but in reality it's none of those three. The problem is exactly what it
seems. The problem is solvable. And yet the smallest solution is:
a = 154476802108746166441951315019919837485664325669565431700026634898253202035277999
b = 36875131794129999827197811565225474825492979968971970996283137471637224634055579
c = 4373612677928697257861252602371390152816537558161613618621437993378423467772036
What the heck is going on??! If you search on the internet, you will
see long-winded explanations about how the problem is actually connected
to elliptic
curves, and other fancy buzzwords from algebraic geometry that
you've never heard of even if you think you know everything about
elliptic curves because you're a cryptographer crypto
enthusiast.
The goal of this post will be to give a maximally elementary
explanation of this puzzle that does not rely on any pre-existing
knowledge of these concepts.
First, let us start off by looking at a relaxed version of
the problem. Specifically, let's remove the requirement that the three
values must be positive. It turns out that you can just try a whole
bunch of possibilities and get two answers by pure human brute force:
\((-11, -4, 1)\) and \((11, -5, 9)\). You can get more solutions
from either of these two by re-arranging the numbers, flipping signs and
multiplying by constant factors (eg. \((-2, 8,
22)\) is also a valid solution), but we'll treat those as being
the same.
Now, we get to the interesting part. What if we can come up with a
way to combine two solutions and get a totally new
third solution? Often, this is possible in mathematics: if you
have two multiples of 5, their sum is also a multiple of 5. More
nontrivially, if you have two rational points on a circle (ie. \((\frac{p_1}{q_1}, \frac{r_2}{s_2})\) and
\((\frac{p_2}{q_2}, \frac{r_2}{s_2})\)
satisfying \((\frac{p_i}{q_i})^2 +
(\frac{r_i}{s_i})^2 = 1\)), you can use point addition laws to
make a third rational point on the circle: \((\frac{p_1 p_2}{q_1 q_2} - \frac{r_1 r_2}{s_1
s_2}, \frac{p_1 r_2}{q_1 s_2} - \frac{r_1 p_2}{s_1 q_2}\)).
If we can come up with such an algorithm for our problem,
then we could just use it over and over again, until eventually we get a
point that happens to be all-positive by pure luck. This will be our
solution path.
Let us simplify the problem by making it two-dimensional. Note that
the equation is homogeneous: it has the property that scaling
all the inputs by the same number does not change the answer. Let's use
this to remove the \(c\) variable:
\(\frac{a}{b+1} + \frac{b}{a+1} +
\frac{1}{a+b} - 4 = 0\)
We also moved the 4 to the left, this will make things more
convenient for us later. Any solution to this with rational
\(a = \frac{p}{q}\) and \(b = \frac{r}{s}\) corresponds to an
integer solution to the original equation: \((a * qs, b * qs, qs)\).
Now, let's multiply by the denominators, to make the whole equation a
pure polynomial:
\(a(a+1)(a+b) + b(b+1)(a+b) + (a+1)(b+1) -
4(a+1)(b+1)(a+b) = 0\)
This introduces a few extraneous solutions, eg. \((a = 1, b = -1)\). But we just ignore them.
We can plot the above expression as a function; it looks like this:
Now, let's draw a line through the two points where we know this
curve intersects the \(z=0\) plane:
\((-11, -4)\) and \((\frac{11}{9}, \frac{-5}{9})\). We derive
these by starting from the two solutions to the three-value equation,
and converting \((a,b,c) \rightarrow
(\frac{a}{c}, \frac{b}{c})\).
And as we can see, the line has a third point of intersection. And
because the line is along the \(z=0\)
plane, the third point must also be along the \(z=0\) plane, ergo it must also be a
solution. Now, let us prove that this point is rational, so it
still corresponds to a valid solution to the original problem.
We can parametrize the line as \((a = -11 +
(\frac{11}{9} + 11) * t, b = -4 + (\frac{-5}{9} + 4) * t)\).
\(t=0\) gives the first point, \(t=1\) gives the second point. Then, we can
look at how \(a(a+1)(a+b) + b(b+1)(a+b) +
(a+1)(b+1) - 4(a+1)(b+1)(a+b)\) behaves along the line
by substituting \(a\) and \(b\) with their expressions based on \(t\).
This gives us the curve: \(y =
\frac{9071}{81} * t^3 + \frac{16748}{81} * t^2 + \frac{7677}{81} *
t\). Here's a plot of that curve, showing all three
intersections:
You can then take the new \(t\) and
plug it into the line and get back your new point: \((\frac{-5951}{9071},
\frac{-9841}{9071})\).
The underlying reason the new \(t\)
(and hence the new point) is rational is: the curve is a degree-3
polynomial in \(t\), and if a degree-3
polynomial has three solutions, then it fully factors into \(k(x-a)(x-b)(x-c)\). The degree-2 term of
this equals \(-k(a+b+c)\). Hence, if
the degree-2 and degree-3 terms are rational (which they are, because we
constructed the polynomial out of an equation with rational
coefficients), and two of the solutions are rational, the third solution
must be rational as well.
From here, we can brute-force our way to getting more solutions.
Unfortunately, we can't directly keep going with the three points we
have: if you were to repeat the above process to generate a new point
starting from any two of the three points we now have, you would just
get back the third point. To get past this, we'll use a clever trick to
generate even more solutions: convert \((x,
y)\) into \((y, x)\).
Now, with this in mind, we just brute-force, repeatedly using the
"find the third point on the line" algorithm and coordinate flipping to
get as many new solutions as possible. Here's the python code:
from sympy import symbols, Poly, solve, simplify, Rational, expand
from math import lcm
def find_third_point(a1, b1, a2, b2):
"""
Find the third intersection point of a line through points (a1, b1) and (a2, b2)
on the cubic curve a(a+1)(a+b) + b(b+1)(a+b) + (a+1)(b+1) = 4(a+1)(b+1)(a+b).
Args:
a1, b1: Coordinates of the first point (rational numbers)
a2, b2: Coordinates of the second point (rational numbers)
Returns:
Tuple (a3, b3): Coordinates of the third intersection point
"""
# Define symbolic variables
t, a, b = symbols('t a b')
# Parameterize the line: a = a1 + t*(a2 - a1), b = b1 + t*(b2 - b1)
a_expr = a1 + t * (a2 - a1)
b_expr = b1 + t * (b2 - b1)
# Define the cubic curve equation:
# a(a+1)(a+b) + b(b+1)(a+b) + (a+1)(b+1) = 4(a+1)(b+1)(a+b)
left_side = a * (a + 1) * (a + b) + b * (b + 1) * (a + b) + (a + 1) * (b + 1)
right_side = 4 * (a + 1) * (b + 1) * (a + b)
equation = left_side - right_side
# Substitute the line parameterization into the equation
poly_t = equation.subs({a: a_expr, b: b_expr})
# Simplify and convert to a polynomial in t
poly_t = expand(poly_t)
poly = Poly(poly_t, t)
# Get the coefficients of the cubic polynomial
coeffs = poly.coeffs()
# The polynomial is cubic (degree 3), so coeffs = [c3, c2, c1, c0]
# For a cubic c3*t3 + c2*t2 + c1*t + c0, the sum of roots is -c2/c3
while len(coeffs) < 4:
coeffs.append(0)
if len(coeffs) != 4:
raise ValueError("Unexpected polynomial degree. Expected cubic polynomial.")
c3, c2, c1, c0 = coeffs
# The known roots are t=0 (for point 1) and t=1 (for point 2)
# Sum of roots: t1 + t2 + t3 = -c2/c3
# Since t1 = 0, t2 = 1, we have 0 + 1 + t3 = -c2/c3
t3 = -c2 / c3 - (0 + 1)
# Compute the third point coordinates
a3 = a1 + t3 * (a2 - a1)
b3 = b1 + t3 * (b2 - b1)
# Simplify the results to ensure rational output
a3 = simplify(a3)
b3 = simplify(b3)
return (a3, b3)
# Verify the a point is on the curve
def is_on_curve(a, b):
left = a * (a + 1) * (a + b) + b * (b + 1) * (a + b) + (a + 1) * (b + 1)
right = 4 * (a + 1) * (b + 1) * (a + b)
return left == right
def is_too_big(x, y):
xn, xd = x.as_numer_denom()
yn, yd = y.as_numer_denom()
return max(abs(xn), abs(xd), abs(yn), abs(yd)) > 10**200
def find_smallest_all_positive_point():
points = [(Rational(-11, 1), Rational(-4, 1)), (Rational(11, 9), Rational(-5, 9))]
queue = [(points[0], points[1])]
def accept(x, y):
print(x, y)
for (x2, y2) in points:
queue.append(((x2, y2), (x,y)))
points.append((x, y))
while len(queue) > 0:
print(len(queue))
(x1, y1), (x2, y2) = queue.pop()
x3, y3 = find_third_point(x1, y1, x2, y2)
if not is_too_big(x3, y3):
if (x3, y3) not in points:
accept(x3, y3)
if (y3, x3) not in points:
accept(y3, x3)
print(points)
eligible = [(x,y) for (x,y) in points if x > 0 and y > 0]
return min(
eligible,
key=lambda x: lcm(x[0].as_numer_denom()[1], x[1].as_numer_denom()[1])
)
This is horribly inefficient, but it does the job. And out we
get:
(154476802108746166441951315019919837485664325669565431700026634898253202035277999/4373612677928697257861252602371390152816537558161613618621437993378423467772036, 36875131794129999827197811565225474825492979968971970996283137471637224634055579/4373612677928697257861252602371390152816537558161613618621437993378423467772036)
This is a positive rational solution to the two-equation formula; add
\(c=1\) and you get a positive rational
solution to the three-equation formula. And from here, we multiply by
the denominator, and get the original solution:
a = 154476802108746166441951315019919837485664325669565431700026634898253202035277999
b = 36875131794129999827197811565225474825492979968971970996283137471637224634055579
c = 4373612677928697257861252602371390152816537558161613618621437993378423467772036
This does not prove that this is the smallest solution: for
that you actually would have to go into the much deeper elliptic curve
theory that I tried hard to avoid. But it gives you a solution to the
problem, and the underlying math actually is elliptic curve math in
disguise: "find the third intersecting point on the line and flip the
coordinates" is exactly the same thing as the elliptic
curve addition law, except flipping across the diagonal axis instead
of the x axis. This "addition law" we came up with even satisfies
associativity.
A simple explanation of a/(b+c) + b/(c+a) + c/(a+b) = 4
2025 May 11 See all postsYou may have at some point seen this math puzzle:
The puzzle has gained some degree of notoriety on the internet because it looks like it must be either simple, impossible, or a trick question (solve P = NP? Hahaha, the answer is N = 1, got you!), but in reality it's none of those three. The problem is exactly what it seems. The problem is solvable. And yet the smallest solution is:
What the heck is going on??! If you search on the internet, you will see long-winded explanations about how the problem is actually connected to elliptic curves, and other fancy buzzwords from algebraic geometry that you've never heard of even if you think you know everything about elliptic curves because you're a
cryptographercrypto enthusiast.The goal of this post will be to give a maximally elementary explanation of this puzzle that does not rely on any pre-existing knowledge of these concepts.
First, let us start off by looking at a relaxed version of the problem. Specifically, let's remove the requirement that the three values must be positive. It turns out that you can just try a whole bunch of possibilities and get two answers by pure human brute force: \((-11, -4, 1)\) and \((11, -5, 9)\). You can get more solutions from either of these two by re-arranging the numbers, flipping signs and multiplying by constant factors (eg. \((-2, 8, 22)\) is also a valid solution), but we'll treat those as being the same.
Now, we get to the interesting part. What if we can come up with a way to combine two solutions and get a totally new third solution? Often, this is possible in mathematics: if you have two multiples of 5, their sum is also a multiple of 5. More nontrivially, if you have two rational points on a circle (ie. \((\frac{p_1}{q_1}, \frac{r_2}{s_2})\) and \((\frac{p_2}{q_2}, \frac{r_2}{s_2})\) satisfying \((\frac{p_i}{q_i})^2 + (\frac{r_i}{s_i})^2 = 1\)), you can use point addition laws to make a third rational point on the circle: \((\frac{p_1 p_2}{q_1 q_2} - \frac{r_1 r_2}{s_1 s_2}, \frac{p_1 r_2}{q_1 s_2} - \frac{r_1 p_2}{s_1 q_2}\)).
If we can come up with such an algorithm for our problem, then we could just use it over and over again, until eventually we get a point that happens to be all-positive by pure luck. This will be our solution path.
Let us simplify the problem by making it two-dimensional. Note that the equation is homogeneous: it has the property that scaling all the inputs by the same number does not change the answer. Let's use this to remove the \(c\) variable:
\(\frac{a}{b+1} + \frac{b}{a+1} + \frac{1}{a+b} - 4 = 0\)
We also moved the 4 to the left, this will make things more convenient for us later. Any solution to this with rational \(a = \frac{p}{q}\) and \(b = \frac{r}{s}\) corresponds to an integer solution to the original equation: \((a * qs, b * qs, qs)\).
Now, let's multiply by the denominators, to make the whole equation a pure polynomial:
\(a(a+1)(a+b) + b(b+1)(a+b) + (a+1)(b+1) - 4(a+1)(b+1)(a+b) = 0\)
This introduces a few extraneous solutions, eg. \((a = 1, b = -1)\). But we just ignore them. We can plot the above expression as a function; it looks like this:
Now, let's draw a line through the two points where we know this curve intersects the \(z=0\) plane: \((-11, -4)\) and \((\frac{11}{9}, \frac{-5}{9})\). We derive these by starting from the two solutions to the three-value equation, and converting \((a,b,c) \rightarrow (\frac{a}{c}, \frac{b}{c})\).
And as we can see, the line has a third point of intersection. And because the line is along the \(z=0\) plane, the third point must also be along the \(z=0\) plane, ergo it must also be a solution. Now, let us prove that this point is rational, so it still corresponds to a valid solution to the original problem.
We can parametrize the line as \((a = -11 + (\frac{11}{9} + 11) * t, b = -4 + (\frac{-5}{9} + 4) * t)\). \(t=0\) gives the first point, \(t=1\) gives the second point. Then, we can look at how \(a(a+1)(a+b) + b(b+1)(a+b) + (a+1)(b+1) - 4(a+1)(b+1)(a+b)\) behaves along the line by substituting \(a\) and \(b\) with their expressions based on \(t\).
This gives us the curve: \(y = \frac{9071}{81} * t^3 + \frac{16748}{81} * t^2 + \frac{7677}{81} * t\). Here's a plot of that curve, showing all three intersections:
You can then take the new \(t\) and plug it into the line and get back your new point: \((\frac{-5951}{9071}, \frac{-9841}{9071})\).
The underlying reason the new \(t\) (and hence the new point) is rational is: the curve is a degree-3 polynomial in \(t\), and if a degree-3 polynomial has three solutions, then it fully factors into \(k(x-a)(x-b)(x-c)\). The degree-2 term of this equals \(-k(a+b+c)\). Hence, if the degree-2 and degree-3 terms are rational (which they are, because we constructed the polynomial out of an equation with rational coefficients), and two of the solutions are rational, the third solution must be rational as well.
From here, we can brute-force our way to getting more solutions. Unfortunately, we can't directly keep going with the three points we have: if you were to repeat the above process to generate a new point starting from any two of the three points we now have, you would just get back the third point. To get past this, we'll use a clever trick to generate even more solutions: convert \((x, y)\) into \((y, x)\).
Now, with this in mind, we just brute-force, repeatedly using the "find the third point on the line" algorithm and coordinate flipping to get as many new solutions as possible. Here's the python code:
This is horribly inefficient, but it does the job. And out we get:
This is a positive rational solution to the two-equation formula; add \(c=1\) and you get a positive rational solution to the three-equation formula. And from here, we multiply by the denominator, and get the original solution:
This does not prove that this is the smallest solution: for that you actually would have to go into the much deeper elliptic curve theory that I tried hard to avoid. But it gives you a solution to the problem, and the underlying math actually is elliptic curve math in disguise: "find the third intersecting point on the line and flip the coordinates" is exactly the same thing as the elliptic curve addition law, except flipping across the diagonal axis instead of the x axis. This "addition law" we came up with even satisfies associativity.