Level 1 Mathematics Olympiad contest! - By Swapnil

1. Prove that 4 does not divide $m^{2}+2$

So, let m be any positive integer. Now 'm' can be written as $m= 4b+r$ where $( 0 \leq r < 4)$ Now this means all the positive integers in the world can be represented in this form.

Now the value of $r = 0,1,2,3$

$1. If$ $r=0$ then $m= 4b+r$

$m= 4b$

$m^{2} = 4b^{2}$

$m^{2} = 16b^{2}$

But it is given that whether 4 is divided by $m^{2}+2$

Now, $m^{2}+2$ = $16b^{2} +2$= $4(4b^{2})+2$

Now, $4(4b^{2})+2 \equiv 2 \pmod{4}$

So, we are getting remainder $2$. So for first case it is not divisible.

$2. If$ $r=1$ then $m= 4b+r$

$m= 4b+1$

$m^{2}= (4b+1)^{2}$

$m^{2} = 16b^{2} + 8b +1$

$m^{2}= 4(4b^{2} + 2b) +1$

But it is given that whether 4 is divided by $m^{2}+2$

Now, $m^{2}+2$ = $4(4b^{2} + 2b) +1+2$

$m^{2}+2$ = $4(4b^{2} + 2b) + 3$

Now, $4(4b^{2}+2b)+3 \equiv 3 \pmod{4}$

So, we are getting remainder $3$. So for second case it is not divisible.

$3. If$ $r=2$ then $m= 4b+r$ $m= 4b+2$

$m^{2}= (4b+2)^{2}$

$m^{2} = 16b^{2} + 16b +4$

$m^{2}= 4(4b^{2} + 4b+1)$

But it is given that whether 4 is divided by $m^{2}+2$

Now, $m^{2}+2$ = $4(4b^{2} + 4b+1) +2$

Now, $4(4b^{2}+4b+1)+2 \equiv 2 \pmod{4}$

So, we are getting remainder $2$. So for third case it is not divisible.

$4. If$ $r=3$ then $m= 4b+r$

$m= 4b+3$

$m^{2}= (4b+3)^{2}$

$m^{2} = 16b^{2} + 24b +9$

$m^{2} = 16b^{2} + 24b +8+1$

$m^{2}= 4(4b^{2} + 6b+2)+1$

But it is given that whether 4 is divided by $m^{2}+2$

Now, $m^{2}+2$ = $4(4b^{2} + 6b+2)+1+2$

$= 4(4b^{2} + 6b+2)+3$

Now, $4(4b^{2}+6b+2)+3 \equiv 3 \pmod{4}$

So, we are getting remainder $3$. So for forth case it is not divisible.

As in all fours cases if $m^{2}+2$ is divided by 4 we are getting some remainders . That means 4 does not divide $m^{2}+2$for any integer m

$The$ Last two digits of $3^{1997}$ is 63. Just see the pattern.

1. For every integer $m$, when it's square is divided by 4 , it will give out 1 or 0 respectively. So when these 2 numbers are each added by 2, it gives out 2 and 3 respectively. What's more, for every integer $m$ which is an even number, its square has at least 2 2's, so for every integer $m$ which is even, it has at least 2×2=$\boxed{4}$. These 2 ways can prove that for every integer $m$, $m^{2}+2$ will never be divisible by 4.

1. Find the last 2 digits of $3^{1997}$

$3^{1}$=3, $3^{2}$=9, $3^{3}$=27, $3^{4}$=81, $3^{5}$=43, $3^{6}$=29, $3^{7}$=87, $3^{8}$=61, $3^{9}$=83, $3^{10}$=49, $3^{11}$=47, $3^{12}$=41, $3^{13}$=23, $3^{14}$=69, $3^{15}$=07, $3^{16}$=21, $3^{17}$=63, $3^{18}$=89, $3^{19}$=67, $3^{20}$=01. And so on......

$\frac{1997}{20}=99......17.$

Refer to $3^{17}$.

So, the last 2 digits of $3^{1997}$ = 63. LoL

It's a VERY VERY old method...... XD

For the 2nd question, my solution is as follows :

3^1997 = (3^6)^332 * 3^5 (mod 100)

             = (29)^332 * 43                 (mod 100)

             =   (29^2)^166 * 43          (mod 100)

             = (41)^166 * 43                  (mod 100)

             = (41^2)^83 * 43               (mod 100)

             = (81)^83 * 43                      (mod 100) 

             = (81^2)^41 * 81 * 43        (mod 100)

             = (61)^41 * 83                       (mod 100)

             = (61^2)^20 * 61 * 83         (mod 100)

             = (21)^20 * 63                        (mod 100)

             = (21^2)^10 * 63                   (mod 100)

             = ( 41)^10 *63                        (mod 100)

              = ( 41^2)^5 * 63                    (mod 100)

              = (81)^5 *63                           (mod 100)

              = (81^2)^2 * 81 *63             (mod 100)

              = (61)^2 *3                               (mod 100) 

              = 21 * 3                                       (mod 100)     

              = 63                                             (mod 100)

Therefore, the last 2 digits of 3^1997 are 63

Hey bros, I live very very far away from you. When would you like to come to my country?

$m^2 + 2 = 2 or 3 (mod4)$

but $4 = 0 (mod4)$

A contradiction.