Find natural numbers m, n where m²-4n and n²-4m are perfect squares

[Albert1]

$m,n \in N$
if $m^2-4n ,\,\, and \,\, n^2-4m$ are all perfect square numbers , please find all $m$ and $n$

 

[Wilmer]

Should you not have some restrictions, like m > n?

Spoiler

Asking because these are 3 of the solutions:
m = 4, n = 4
m = 6, n = 5
m = 5, n = 6

 

[Albert1]

Should you not have some restrictions, like m > n?

Spoiler

Asking because these are 3 of the solutions:
m = 4, n = 4
m = 6, n = 5
m = 5, n = 6

your answers are corret,please state with reason

 

[kaliprasad]

Spoiler

without loss of generality we can assume m >=n. if(m,n) is a solution then (n,m) is another
now 4n is even so $m^2 - 4n <= (m-2)^2$
or $-4n<=-4m + 4$
or $n >=m-1$

case 1)
$m=n$
then $m^2 -4m = m(m-4)$ is perfect square = $(k+2)(k-2)$ taking $k = m-2$

$(k+2)(k-2) = k^2 - 4 = p^2(say)$
$(k+p)(k-p) = 4$ so $k+ p = k- p = 2$ as both are even or both odd
so $k = 2, p = 0$ or $m = 4$
so solution $m=n= 4$

case 2)
$n >=m-1$
or $n = m- 1$
now $m^2- 4n = m^2-4(m-1) = (m-2)^2$ always whole square
check the other one
$n^2 - 4m = n^2 - 4(n+1) = n^2 - 4n - 4 = (n-2)^2 + 8 = q^2(say)$

or $(n+2)^2-q^2 = 8$ giving $n-2 + q =4$ and $n-2-q=2$ as both are even
solving $n= 5$ and as $m = n+1$ we have (m,n) = (6,5)
by symmetry (m,n) = (5,6)

combining both we have (m,n) = (6,5) or (5,6) or (4,4)

 

[Wilmer]

your answers are correct,please state with reason

Sorry. Was intending to ask question only...gave those as examples.

Didn't realize negative integers were not allowed.
If they were, then there's infinite solutions:
if m = x then n = -(x + 1)

 

[Albert1]

Sorry. Was intending to ask question only...gave those as examples.

Didn't realize negative integers were not allowed.
If they were, then there's infinite solutions:
if m = x then n = -(x + 1)

yes this original qustion has a flaw:
here we treat $m^2-4n , and \, n^2-4m $ separately