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 $n=7k\ or\ 7k-2 \ or\ 7k+2$ )

#NumberTheory #Divisor #PerfectSquare #Multiples

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Comments

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 - 7 years, 4 months ago

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

Megh Parikh - 7 years, 4 months ago

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 - 7 years, 4 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}`	$\boxed{123}$