let $m,n \in N$ such that $2m^2+m=3n^2+n$, Find all integral solution of $2m^2+m=3n^2+n$?

Question

let $m,n \in N$ such that $2m^2+m=3n^2+n$. Find all integral solution of $2m^2+m=3n^2+n$ ?

My work:

$$2m^2-2n^2+m-n=n^2$$

$$\implies 2(m-n)(m+n)+(m-n)=n^2$$

$$\implies(2m+2n+1)(m-n)=n^2$$

Case $1$

$2m+2n+1=n^2$ and $m-n=1$

Now putting $m=n+1$ in $2m+2n+1=n^2$

which lead me to $2(n+1)+2n+1=n^2$

$$\implies n^2-4n-3=0$$

above equation give me irrational roots.

Case 2

$2m+2n+1=n$ and $m-n=n$

Now putting $m=2n$ in $2m+2n+1=n$

which leads me to $2(2n)+2n+1=n$

$\implies n=\frac{-1}{5}$

which is not a natural number.

I am not getting any solution to the given equations. Is my approach correct?

Also is it necessary that $(m-n)$ and $(2m+2n+1)$ must be prefect square

Please Suggest a solution without using Modular Arithmetic

Why aren’t other cases possible? $n^2$ might have more factors. — Michael Burr, May 28 '21 at 11:17
Instead of saying $x \equiv 0 \bmod 3$ one can use $3\mid x$. I think you cannot avoid such arithmetic. — Dietrich Burde, May 28 '21 at 11:30
https://artofproblemsolving.com/community/c3046h1055724_almost_pell — individ, May 29 '21 at 05:27

Sathvik · Accepted Answer · 2021-05-28T12:22:14.110

4

$$\begin{align*} 2m^2+m&=3n^2+n \\ \frac{1}{8}\left((4m+1)^2-1\right)&=\frac{1}{12}\left((6n+1)^2-1\right)\\ 3(4m+1)^2-3&=2(6n+1)^2-2 \end{align*}$$ Let $x=(4m+1)^2$ and $y=(6n+1)^2$, we have, $$3x^2-2y^2=1$$ The above equation has been solved here

The general solution is obtained from the recurrences, $$x_{n+1}=5x_n+4y_n,\;y_{n+1}=6x_n+5y_n,\;(x_1,y_1)=(1,1). $$

edited May 28 '21 at 12:22

answered May 28 '21 at 12:01

Sathvik

3,609

how will i prove that both $m-n$ and $2m+2n+1$ are perfect square? – mathophile May 28 '21 at 13:18
1

Hint: Multiply the two expressions and think about $\gcd$. – Sathvik May 28 '21 at 13:57
$gcd((m-n),(2m+2n+1))=1$ only if $\gcd(m,n)=1$. But it is not given that $\gcd(m,n)=1$ So how to find $\gcd((m-n),(2m+2n+1))$ – mathophile Jun 01 '21 at 08:55
Hint: Prove the following: $m-n=\gcd(m^2,n^2)=\gcd(m,n)^2$. The required result will follow. – Sathvik Jun 01 '21 at 09:26
@Sathvik can u please elaborate it further ?? – Ash_Blanc Aug 06 '23 at 14:41

Will Jagy · Answer 2 · 2021-05-28T20:23:26.623

one may resolve the combined recurrence into separate degree two linear recurrences

$$ m_{k+2} = 98 \; m_{k+1} - m_k + 24 $$ with the sequence beginning $0, 22, 2180, 213642, ... $

$$ n_{k+2} = 98 \; n_{k+1} - n_k + 16 $$ with the sequence beginning $0, 18, 1780, 174438, ... $

The common values $v= 2 m^2 + m = 3 n^2 + n$ begin $$ 0, 990, 9506980, 91286021970, 876528373449960, 8416425350580494950, $$

and obey $$ v_{k+2} = 9602 \; v_{k+1} - v_k + 1000 $$

There was a question about $d_k = m_k - n_k.$ This is a square, $d_k$ beginning $0, 4, 400,...$ and $$ d_{k+2} = 98 \; d_{k+1} - d_k + 8 $$ If we name $s_k = \sqrt{m_k - n_k}$ we find that $s_k$ are integers, beginning $0, 2, 20, 198,...$ and obeying $$ s_{k+2} = 10 \; s_{k+1} - s_k $$

let $m,n \in N$ such that $2m^2+m=3n^2+n$, Find all integral solution of $2m^2+m=3n^2+n$?

2 Answers2