# Thread: modified pythagorus theorem!

1. ## modified pythagorus theorem!

Hello,

I need a little help in figuring this weird problem...

--we know that (2+3)^2 is not equal to 2^2 and 3^2...
however... (2+3)^2 = (2+1)^2 + (3+1)^2!

so determine, all values for m and n such that---
(m+n)^2 = (m+1)^2 + (n+1)^2

Thanks!

2. Originally Posted by Vedicmaths
Hello,

I need a little help in figuring this weird problem...

--we know that (2+3)^2 is not equal to 2^2 and 3^2...
however... (2+3)^2 = (2+1)^2 + (3+1)^2!

so determine, all values for m and n such that---
(m+n)^2 = (m+1)^2 + (n+1)^2

Thanks!
$(m+n)^2 = (m+1)^2 + (n+1)^2$
expand & simplify results in
$mn = m+n+1$
If the product of any two numbers is equal to the sum of the two numbers plus 1, then they work.

Is that what you had in mind?
2,3 is a solution that you have given.
It appears that is the only solution in natural numbers.
.

3. Originally Posted by Vedicmaths
Hello,

I need a little help in figuring this weird problem...

--we know that (2+3)^2 is not equal to 2^2 and 3^2...
however... (2+3)^2 = (2+1)^2 + (3+1)^2!

so determine, all values for m and n such that---
(m+n)^2 = (m+1)^2 + (n+1)^2

Thanks!
Hmmm....isn't this trivial? Just develop parentheses in both sides!

m^2 + 2mn + n^2 = m^2 + 2m + 1 + n^2 + 2n + 1 ==> 2mn = 2(m+n+1) <==> mn - m - n = 1 <==> (m-1)(n-1) = 2, so we don't have that many options for m, n , do we?

Tonio

4. thanks for the response!

yeah I also simplified that and ended up getting (m-1)(n-1) = 2...that means m and n can only have 2 or 3 values respectively.
I was just making it too complicated..I thought it would be a little bit harder that it seems!

Hence, m =2 or 3 and n = 3 or 2...right?

Thanks a lot!

5. Originally Posted by Vedicmaths
thanks for the response!

yeah I also simplified that and ended up getting (m-1)(n-1) = 2...that means m and n can only have 2 or 3 values respectively.
I was just making it too complicated..I thought it would be a little bit harder that it seems!

Hence, m =2 or 3 and n = 3 or 2...right?

Thanks a lot!

Yep. Since the equation (m-1)(n-1) = 2 is symmetric wrt m, n there's only one solution in the naturals: m = 3, n = 2.
If you allow integers in general there's another solution: m = 0, n = -1

Tonio