×

# Proof Problem

Find all the integral solutions of $${ x }^{ 4 }+{ y }^{ 4 }+{ z }^{ 4 }-{ w }^{ 4 }=1995$$.

Note by Swapnil Das
2 years, 5 months ago

MarkdownAppears as
*italics* or _italics_ italics
**bold** or __bold__ bold
- bulleted- list
• bulleted
• list
1. numbered2. list
1. numbered
2. list
Note: you must add a full line of space before and after lists for them to show up correctly
paragraph 1paragraph 2

paragraph 1

paragraph 2

[example link](https://brilliant.org)example link
> This is a quote
This is a quote
    # I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
# I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
MathAppears as
Remember to wrap math in $$...$$ or $...$ to ensure proper formatting.
2 \times 3 $$2 \times 3$$
2^{34} $$2^{34}$$
a_{i-1} $$a_{i-1}$$
\frac{2}{3} $$\frac{2}{3}$$
\sqrt{2} $$\sqrt{2}$$
\sum_{i=1}^3 $$\sum_{i=1}^3$$
\sin \theta $$\sin \theta$$
\boxed{123} $$\boxed{123}$$

Sort by:

We know that $$a^4 \equiv 0,1 \pmod {16} \quad \forall \ a \in \mathbb Z$$.

We can have:

$x^4 + y^4 + z^4 - w^4 \equiv -1,0,1,2,3 \pmod {16}$

or

$x^4 + y^4 + z^4 - w^4 \equiv 0,1,2,3,15 \pmod {16}$

But $$1995 \equiv 11 \pmod {16}$$. So no integral solutions exist!

@Swapnil Das - I'd say it's a Number Theory Problem!

- 2 years, 5 months ago

Solution master: @Satyajit Mohanty

- 2 years, 5 months ago

Yeah! It is! Great solution @Satyajit Mohanty :)

- 2 years, 5 months ago

why did you work with mod16? why not other numbers? is there any rule?

- 2 years, 5 months ago

I worked with mod 16 because fourth powers have an interesting property as they are equivalent to only 0 or 1 modulo 16. I don't have any idea about other modulos for 16th powers. Well, one can try further.

- 2 years, 5 months ago