Question

if m and n are integers, find values of m and n such that m^2 and n^2 - 12

Answer 1

... they what? .-.

Answer 2

what do you mean

Answer 3

if m and n are integers, find values of m and n such that m^2 and n^2 - 12 .... what should they equal, ... or what condition should they satisfy?

Answer 4

they should both equal 21
when n(squared) is subtracted from m(squared)

Answer 5

?

Answer 6

They can't equal \(21\).
They gotta be \(42\).

Answer 7

so you want m^2 + n^2 = 21?

Answer 8

how
an example of such a solution is
\[5^{2}-2^{2}=21\]

Answer 9

So\[(m +n)(m - n) = 21\]

Answer 10

Diophantine...

Answer 11

exactly

Answer 12

\(n+m=7,n-m=3\) solve for \(n,m\)

Answer 13

since \(21=7\times 3\)

Answer 14

Right there.

Answer 15

\[n - m = 3, n + m = 7\]

Answer 16

no

Answer 17

also \(m+n=21,m-n=1\) will give another solution

Answer 18

Oh I mean\[n + m = 3 , n - m = 3\]

Answer 19

nope
it is a quadratic expression

Answer 20

i wish i knew what the original question actually said

Answer 21

and it is m^2-n^2=21

Answer 22

i just need numbers that make 21 when they are subtracted

Answer 23

\[m^2 - n^2 = (m + n)(m - n)\]Proceed.

Answer 24

Wait, numbers when they are subtracted?\[m -n = 21\]?

Answer 25

the squares

Answer 26

got it
then what @ParthKohli said will work. solve the system (in your head)
\[m+n=7\]\[m-n=3\]

Answer 27

Eliminate, eliminate, eliminate, pirate!

Answer 28

\(m=5,n=2\)

Answer 29

that is one

Answer 30

i need one more

Answer 31

or \[m+n=21\]\[m-n=1\]
\[m=11,m=10\]

Answer 32

\[m - n = 21 \\ m +n = 1\]

Answer 33

in other words there are two ways to factor \(21\) either use \(7\times 3\) or \(21\times 1\)

Answer 34

@parthkohli i agree that it is 7 and 3 but i already got that i need one more example

Answer 35

since \(m^2-n^2=(m+n)(m-n)\) you use either factorization

Answer 36

Actually if you look,\[m = \pm 21, n = \pm 1\]

Answer 37

wait i got it
its \[\pm10\] and \[\pm11\]