Fred Agbo & Jed
September 14, 2026
hVSGho
A Pythagorean triplet is a set of three natural numbers, \(a < b < c\), for which, \[ a^2 + b^2 = c^2\]
There exists exactly one Pythagorean triplet where \(a + b + c = 1000\). Find it.
# My own attempt below!
def pythagorean(a,b,c):
"""
Identify the contraints
1. The Ordering: a < b < c (This prevents duplicate searches, like checking both a=3, b=4 and a=4, b=3).
2. The Equation: a^2 + b^2 = c^2 (The Pythagorean theorem).
3. The Sum: a + b + c = 1000.
Reduce the Variables (Mathematical Optimization)
1. Since a + b + c = 1000, you can rewrite c as: c=1000-a-b
2. In Pythagorean a^2 + b^2 = (1000 - a - b)^2
Establish Strict Loop Boundaries
1. Boundaries for a: Since a < b < c, a must be strictly less than a third of the total sum.
Max a < 1000/3 => a <=332
2. Boundaries for b: b must always start at a + 1 (since a < b).
It must also be less than the remaining sum divided by 2.
Max b < (1000 - a) / 2
ALGORITHMS
Algorithmic Logic With the math simplified, the programmatic blueprint looks like this:
1. Initialize a loop for a ranging from 1 to 332.
2. Initialize a nested loop for b ranging from (a + 1) to (1000 - a) / 2.
3. Calculate c inside the loop: c = 1000 - a - b.
4. Test the condition: Check if a^2 + b^2 == c^2.
Return the result: If the condition is met,
multiply a x b x c to get the final answer and break the loop immediately (since the problem states there is exactly one solution).
"""
# Now Write the program
Discordrich installed easily, I have
included a Python file in the contents that you should just need to run,
and it will (in theory) handle things automatically for you.
p1.py file is doing some stress
testing, but has some issuesp2.py file should
be saying that all systems are a go, but it is notp3.py file should
also be saying that all systems are go, but is notp4.py should be verifying a
scan, but is instead erring out. Use the printed error alongside print
statements or the debugger to establish where things have gone awry