Question

A logical problem (I would type the question)

Answer 1

Show that there exists no natural numbers M and N such that \[m ^{2}=n ^{2}+2002\]

Answer 2

@Hero @thomaster

Answer 3

@ganeshie8

Answer 4

I don't know if i am approaching the answer but i tried to write it as
(m+n)(m-n)=2002

Answer 5

you're on right track, but you will get multiple cases that way

Answer 6

its a one line proof if u know Fermat thm

Answer 7

Fermat little theorem

Answer 8

No but i would like to know

Answer 9

Would you tell me his theorem

Answer 10

Fermat little thm :
\(a^p \equiv a \mod p\)

http://mathworld.wolfram.com/FermatsLittleTheorem.html

Answer 11

How do i apply it here

Answer 12

my internet has problems i will come back after half hour

Answer 13

scratch that Fermat is of no use here, we will do it using the method u started already :

\((m+n)(m-n)=2002 \)

\((m+n)(m-n)=2\times 7 \times 11 \times 13\)

Answer 14

since you have only one "even" prime factor in 2012, it has to go in either (m+n) factor or (m-n) factor, but not both :

\(m+n = 2a\)
\(m-n = 2b+1\)
---------------------
\(2m = 2a + 2b + 1\)

which is impossible as left side is even, but the right side is always odd. QED.

Answer 15

Yes so no sollution to this problem