Interesting Problem!

Prove that : $3^{4^{5}} + 4^{5^{6}} = a.b$

Which a,b is positive integer and each have atleast 2015 digits

#NumberTheory

This discussion board is a place to discuss our Daily Challenges and the math and science related to those challenges. Explanations are more than just a solution — they should explain the steps and thinking strategies that you used to obtain the solution. Comments should further the discussion of math and science.

When posting on Brilliant:

Use the emojis to react to an explanation, whether you're congratulating a job well done , or just really confused .
Ask specific questions about the challenge or the steps in somebody's explanation. Well-posed questions can add a lot to the discussion, but posting "I don't understand!" doesn't help anyone.
Try to contribute something new to the discussion, whether it is an extension, generalization or other idea related to the challenge.
Stay on topic — we're all here to learn more about math and science, not to hear about your favorite get-rich-quick scheme or current world events.

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

Math	Appears 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}`	$\boxed{123}$

Comments

Hint :- $x^4 + 4y^4 = (x^2 + 2y^2 - 2xy)(x^2 + 2y^2 + 2xy)$

Siddhartha Srivastava - 5 years, 9 months ago

Sophie Germain to the rescue!

Calvin Lin Staff - 5 years, 9 months ago

3^4 = 1 (mod 10)
3^4^2 = 1 (mod 10)
.
.
.
3^4^5 = 1 (mod 10)

4^5 = 4 (mod 10)
4^5^2 = 4^5 (mod 10) = 4 (mod 10)
.
.
.
4^5^6  = 4 (mod 10)

3^4^5 + 4^5^6 = (1+4) (mod 10) = 5 (mod 10)

Last digit is 5 so 3^4^5 + 4^5^6 is not prime

Vincent Miller Moral - 5 years, 9 months ago

Be careful, that isn't the way that tower of exponents work. $3^ { 4 ^ 2 } = 3 ^ {16 } = \left( 3^4 \right) ^ 4$ . However, the series of modular arithmetic statements that you wrote are still true.

Note that you are supposed to show that they are the product of 2 digits with > 2015 digits. You have shown that it is a multiple of 5, but that is not sufficient to show that $a, b$ exist.

Calvin Lin Staff - 5 years, 9 months ago

I'm a newbie in Number Theory. Thanks.

Vincent Miller Moral - 5 years, 9 months ago

Math	Appears 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}`	$\boxed{123}$