# No Integer Solutions

Prove that 3x^10 - y^10 = 1991 has no integer solutions.

I was hoping someone could give me a lead on how to approach this question.

Any help is much appreciated!

4 years, 6 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:

Consider the equation modulo $$11$$.

Since $$11$$ divides $$1991$$ it must be that:

$3x^{10}-y^{10} \equiv 0 \pmod{11}$

By Fermat's Little Theorem: $$a^{10} \equiv 1\pmod{11}$$, for integer $$a$$.

Hence, $$3x^{10}-y^{10} \equiv 3-1 \equiv 2 \pmod{11}$$.

Hence no solutions can exist.

- 4 years, 6 months ago

FLT only works for integers coprime to $$11$$, so the other case you need to handle is the possibility that both $$x$$ and $$y$$ are multiples of $$11$$. But this would imply that $$11^{10}$$ divides $$1991$$, which is not the case.

- 4 years, 6 months ago

- 4 years, 6 months ago

Shouldn't you correct the solution?

- 3 years, 6 months ago