Finding Perfect Square Combinations: $m,n\in N$

[Albert1]

$m,n\in N ,n<m $
given :
$(1)1000\leq m<2011$
$(2) m-n=p^k$ here $p$ is a prime, and $k$$\in\{0,1,2\}$
$(3)m\times n $ is a perfect square number , find all possibe $m$

 

[kaliprasad]

$m,n\in N ,n<m $
given :
$(1)1000\leq m<2011$
$(2) m-n=p^k$ here $p$ is a prime, and $k$$\in\{0,1,2\}$
$(3)m\times n $ is a perfect square number , find all possibe $m$

Spoiler

k cannot be zero. because

$x(x+1)$ for integer x >0 is between $x^2$ and $(x+1)^2$ and not perfect square
p cannot be 2 because then we have
$x(x+2)= (x+1)^2-1$ not a perfect square
$x(x+4) = (x+2)^2-4$ not a perfect square
so we have p is odd and now for power 1

$m$ and $m-p$ one is odd and another is even and both have to be perfect square
if not then $m = xa^2$ and $n = xb^2$ and $m-n = x(a^2-b^2)$ not a prime unless x = 1
so both are perfect square.
m and m-p have to be square of consecutive numbers else a^2-b^2 = (a-b)(a+b) is not prime
so m is value x then $2x-1$ is prime
$1000 < 32^2$ and $2000 > 44^2$
so we look for 2 * 32-1 to 2 * 44 -1 to be prime.
they are $67(m= 34^2),71(m=36^2), 73(m= 37^2),79(m= 40^2),83(m=44^2)$

now take the case of $p^2$
there are 2 cases
m and n both are consecutive squares
$p^2$ is between 61 and 88
but there is no p

there are 2nd case that is p is factor of m and n
$n = a^2p$ and $m=b^2p$
and $b^2-a^2$ is divisible by p.
we see by checking the value $m=1900$

 

[Albert1]

Spoiler

k cannot be zero. because

$x(x+1)$ for integer x >0 is between $x^2$ and $(x+1)^2$ and not perfect square
p cannot be 2 because then we have
$x(x+2)= (x+1)^2-1$ not a perfect square
$x(x+4) = (x+2)^2-4$ not a perfect square
so we have p is odd and now for power 1

$m$ and $m-p$ one is odd and another is even and both have to be perfect square
if not then $m = xa^2$ and $n = xb^2$ and $m-n = x(a^2-b^2)$ not a prime unless x = 1
so both are perfect square.
m and m-p have to be square of consecutive numbers else a^2-b^2 = (a-b)(a+b) is not prime
so m is value x then $2x-1$ is prime
$1000 < 32^2$ and $2000 > 44^2$
so we look for 2 * 32-1 to 2 * 44 -1 to be prime.
they are $67(m= 34^2),71(m=36^2), 73(m= 37^2),79(m= 40^2),83(m=44^2)$

now take the case of $p^2$
there are 2 cases
m and n both are consecutive squares
$p^2$ is between 61 and 88
but there is no p

there are 2nd case that is p is factor of m and n
$n = a^2p$ and $m=b^2p$
and $b^2-a^2$ is divisible by p.
we see by checking the value $m=1900$

Spoiler

m=1156,1296,1369,1600,1764 (k=1)
m=1377,1900 (k=2)
you miss 1377

 

[kaliprasad]

Spoiler

m=1156,1296,1369,1600,1764 (k=1)
m=1377,1900 (k=2)
you miss 1377

right

Spoiler

I missed $1377 = 17 * 81 = 17 * 9^2= 17 * (\frac{17+1}{2})^2$
I got $1900 = 19 * 100 = 19 * (\frac{19+1}{2})^2$

basis for solution

Spoiler

for the case k= 2 we have
$m= pa^2$ and $n=n-p^2= pb^2$
and $a^2-b^2=(a-b)(a+b)$ is p so
a-b = 1
a+b = p
or
$m= p(\frac{p+1}{2})^2$
as an estmiate
now m = 1000 gives $p > 15$
m = 2011 given $p < 21$
so p = 17 or 19 and hence the result