MHB Find Perfect Square $m,n \in N$ with $m^2-4n$ & $n^2-4m$

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The discussion focuses on finding natural numbers \( m \) and \( n \) such that both \( m^2 - 4n \) and \( n^2 - 4m \) are perfect squares. Participants highlight the need for restrictions, such as \( m > n \), to narrow down potential solutions. There is acknowledgment of a flaw in the original question, particularly regarding the treatment of the equations separately. The conversation also notes that if negative integers were allowed, there would be infinite solutions. Overall, the emphasis is on clarifying the conditions necessary for solving the problem effectively.
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$m,n \in N$
if $m^2-4n ,\,\, and \,\, n^2-4m$ are all perfect square numbers , please find all $m$ and $n$
 
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Should you not have some restrictions, like m > n?
Asking because these are 3 of the solutions:
m = 4, n = 4
m = 6, n = 5
m = 5, n = 6
 
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Wilmer said:
Should you not have some restrictions, like m > n?
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
 
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)
 
Albert said:
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)
 
Wilmer said:
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
 
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