×

# Positive integer solutions

How many triple of positive integer $$(a,b,c)$$ that satisfy equation: $\frac{c}{17}=\frac{8}{a^2}+\frac{45}{b^2}$

Note by Idham Muqoddas
4 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:

Rearranging the equation we get: $c=\frac{17\cdot 2^3}{a^2}+\frac{17 \cdot 3^2 \cdot 5}{b^2}$ Because the LHS is an integer, the RHS must also be an integer. First, let's address the case in which both fractions in the RHS are integers. Then $$a=1, 2$$ and $$b=1, 3$$, giving us $$4$$ solutions so far.

Now note that when $$a > 11$$ and $$b > 27$$, the RHS is less than one, and thus cannot be a positive integer. You could easily exhaust all 297 possibilities, although I'm sure there is a better way.

- 4 years, 5 months ago