Number theory problem

Both \(4n+1\) and \(3n+1\) are perfect squares.

Then prove that \(56|n\).

From "An Excursion in Mathematics" . (Solution is not given/Challenge Problem)

Pls type your solution in comments section.

(I could prove easily that \(8|n\) and \(n=7k\ or\ 7k-2 \ or\ 7k+2\))

Note by Megh Parikh
4 years, 5 months ago

No vote yet
1 vote

  Easy Math Editor

MarkdownAppears as
*italics* or _italics_ italics
**bold** or __bold__ bold

- bulleted
- list

  • bulleted
  • list

1. numbered
2. 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 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"
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} \( \boxed{123} \)

Comments

Sort by:

Top Newest

If \( 4n+1 = a^2 \) and \( 3n+1 = b^2 \) then \( (2b)^2-3a^2 = 1 \). The solutions to Pell's equation are well known, and \( b \) is even in only every other one. Basically you look at powers of the "fundamental unit" \( 2 + \sqrt{3} \) and set \( 2b \) equal to the first summand and \( a \sqrt{3} \) equal to the second one. For the first term to be even, you need the odd powers of \( 2 + \sqrt{3} \) only.

This leads to the recurrence relation \( a_1 = b_1 = 1 \), \( a_{k+1} = 7a_k + 8b_k, b_{n+1} = 6a_k + 7b_k \). After that it's an easy induction to see that \( a_k \) and \( b_k \) are always \( \pm 1 \) mod \( 7 \), so \( n = (b^2-1)/3 \) should be divisible by \( 7 \).

This is by no means the easiest or best solution, but I thought I should mention that one can generate all the values of \( n \) satisfying the equations: \( 0, 56, 10920, \ldots \).

Patrick Corn - 4 years, 5 months ago

Log in to reply

Problem solution by basic method.I don't know Pellsequation :) Thanks for the help

Megh Parikh - 4 years, 5 months ago

Log in to reply

My half solution:

\(4n+1=a^2\) is \(odd\).

Therefore \(n=2k\) as any perfect square is of form \(8k +1\ OR\ 4k\)

Then \(3n+1\) is also odd.

therefore \(3n+1=8k+1 \implies n=8k\)

for 7 see remainders mod 7 and \[a^2+b^2\equiv2\ (mod\ 7) \] then see \[a^2-b^2\]

Megh Parikh - 4 years, 5 months ago

Log in to reply

×

Problem Loading...

Note Loading...

Set Loading...