diophantine sum of 6 squares

Art of Problem Solving

AoPS Online

High School Olympiads Regional, national, and international math olympiads

Regional, national, and international math olympiads

3 M G

BBookmark VNew Topic kLocked

High School Olympiads Regional, national, and international math olympiads

Regional, national, and international math olympiads

3 M G

BBookmark VNew Topic kLocked

No tags match your search

M

diophantine sum of 6 squares

G H J

Reveal topic tags

Diophantine equation

L

G H BBookmark kLocked kLocked NReply

Source: 2024 RELSMO 3

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

294 posts

#1 h Jun 28, 2024, 3:17 AM

Y by

Find all solutions to
$(abcde)^2 = a^2+b^2+c^2+d^2+e^2+f^2.$ in integers.

Proposed by Seongjin Shim

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

217 posts

#2 h Jun 30, 2024, 3:52 AM • 10 Y

Y by YaoAOPS, yofro, Seungjun_Lee, ATGY, dgkim, avisioner, sixoneeight, peace09, MS_asdfgzxcvb, alexanderhamilton124

My problem!

solution

This post has been edited 3 times. Last edited by mathlove_13520, Jun 30, 2024, 4:03 AM
Reason: Solution hidden and compressed

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

26 posts

zuat.e

#3 h Mar 10, 2025, 7:05 PM

Y by

Only $(a,b,c,d,e)=(0,0,0,0,0,0)$ works.
Suppose no variable is $0$ , then as $n^2-(n-1)^2=2n-1$ , we have $a^2+b^2+c^2+d^2+e^2\geq 2abcde -1$ , from where $c=d=e=1$ , so $(a^2-1)(b^2-1)=f^2+4$ .
Now, if $a^2-1$ or $b^2-1$ are $0\pmod{3}$ , $f^2\equiv2\pmod{3}$ , which is impossible, so $a\equiv b\equiv0\pmod{3}$ , yielding $f\equiv 0\pmod{3}$ , and we finish by infinite descent.

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

217 posts

#4 h Mar 24, 2025, 7:57 AM

Y by

zuat.e wrote:

Only $(a,b,c,d,e)=(0,0,0,0,0,0)$ works.
Suppose no variable is $0$ , then as $n^2-(n-1)^2=2n-1$ , we have $a^2+b^2+c^2+d^2+e^2\geq 2abcde -1$ , from where $c=d=e=1$ , so $(a^2-1)(b^2-1)=f^2+4$ .
Now, if $a^2-1$ or $b^2-1$ are $0\pmod{3}$ , $f^2\equiv2\pmod{3}$ , which is impossible, so $a\equiv b\equiv0\pmod{3}$ , yielding $f\equiv 0\pmod{3}$ , and we finish by infinite descent.

I do not understand why $c=d=e=1$ .
For example, $a=143400000000, b=1434, c=143, d=14, e=1$ works. Could you explain it?

Z K Y

N Quick Reply

G

H

=

© 2025 AoPS Incorporated

a