5) If $$a, b, c, d, e$$ are natural numbers, then prove that $$a^2 + b^3 + c^4 + d^5 = e^6$$ has infinite many solutions.

Note by Dev Sharma
2 years, 7 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:

a solution is (2,3,1,2,2). ssince there exists a solution, we multiply both sides by $$2^{60}$$ which gives us $(2^{30}a)^2+(2^{20}b)^3+(2^{15}c)^4+(2^{12}d)^5=(2^{10}e)^6$ we can repeat this process infinitly giving us infinite solutions. hence solved.

- 2 years, 7 months ago

Nice

- 2 years, 7 months ago

@Aareyan Manzoor can you give the primitive solution!

- 2 years, 7 months ago

(2,3,1,2,2) @Sualeh Asif

- 2 years, 7 months ago

i tried but couldnt find one. let me write a code, i will have a pair in a while...

- 2 years, 7 months ago

$22^2+1^3+1^4+3^5=3^6$ Hence (22,1,1,3,3) is another solution

- 2 years, 6 months ago