Number Theory or Algebra? Part 5

$a^{4} \equiv 0 (mod 4)$ , when $a$ is an even positive integer.

$a^{4} \equiv 1 (mod 4)$ , when $a$ is an odd positive integer.

(The two identities above can be proved by representing all positive integers in form of $2k$ and $2k + 1$ )

$1^{4} + 2^{4} + 3^{4} + ... + 2547^{4} \equiv 1278 (mod 4)$

$\boxed{1^{4} + 2^{4} + 3^{4} + ... + 2547^{4} \equiv \boxed{2} (mod 4)}$

Python code:

s=0

for i in range(2547):

k=i+1

for j in range (4):

    k=k%4

    if j==3:

        k=k

    else:

        k=k*(i+1)



s=s+k

print s%4

Note that even square numbers are 0 (mod 4) and odd square numbers are 1 (mod 4).

So in our series, we just count the number of odd numbers, which is 1274 and calculate 1274 (mod 4), which is 2.

${ a }^{ 4 }\equiv { \left( a\quad \bmod \text{ 4} \right) }^{ 4 }\quad \left( \bmod \text{ 4} \right)$

Now the numbers can be grouped

$636\left( { 1 }^{ 4 }+{ 2 }^{ 4 }+{ 3 }^{ 4 }+{ 4 }^{ 4 } \right) +{ 1 }^{ 4 }+{ 2 }^{ 4 }+{ 3 }^{ 4 }$

Taking modulo 4, the answer is 2.

Just group the numbers in pairs from start and the end ..all the pairs which are of the form a^4 + b^4 will be divisible by 4...in the end the central no. (2547+1)/2 = (1274)^4 will be left..which when divided by 4 leaves a remainder of 2. Hence 2 is the answer.

1^{4}mod4=1 2^{4}mod4=1 i.e (even)^{4}mod4=0 and (odd)^{4}mod4=1
the no. of terms in 1^{4}+2^{4}+.....2547^{4} is 2547 therefore toal no. of one or odd is 2547/2 +1=1274 and the last 2 digit is 74 when we divide 74 by 4 then remainder is 2 therefore answer will be 2