 # number theory

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• #21702

Given,

$a^2+b^2+c^2+d^2=2^{2018}$

So, solve for all possible values of a,b,c,d ?

#21789

SOLUTIONS ARE (0,0,0,2^1009),(2^1008,2^1008,2^1008,2^1008)

ANY SQUARE OF ODD IS CONGRUENT TO 1(MOD 8),SO

THE EQUATION HAS NO SOLUTION WITH AN ODD COMPONENT.

WE MUST HAVE A=2X,B=2Y,C=2Z,D=2W.PUTTING THESE VALUES WE GET

X^2+Y^2+Z^2+W^2=2^2016…CONTINUE THIS PROCESS

WE CAN PROCEED RECURSIVELY AS LONG AS RIGHT HAND SIDE IS 0(MOD8).

EVENTUALLY WE WILL ARRIVE AT L^2+M^2+N^2+P^2=4..

SOLUTIONS OF THIS EQUATION IS (1,1,1,1),(0,0,0,2)…

PUTTING THOSE VALUES WE WILL GET SOLUTIONS..

#21800

Well done @Arpan, the solution is correct. Let me add some detailed proof and give it a beautiful look –

Solutions are $(0,0,0,2^{1009}),(2^{1008},2^{1008},2^{1008},2^{1008})$

Any square of odd number is$\equiv 1(mod 8)$,So the equation has no solution with an odd component, means all of $(a,b,c,d)$ need to be even. So,we must have $a=2x,b=2y,c=2z,d=2w$ . Putting these values we get

$x^2+y^2+z^2+w^2=2^{2016}$ …continuing this process we can proceed as long as we have right hand side $\equiv 0\pmod{8}$.

Eventually we will arrive at $l^2+m^2+n^2+p^2=4$ and solutions to this equation are  $(1,1,1,1),(0,0,0,2)$.

Putting these values we are getting $(0,0,0,2^{1009}),(2^{1008},2^{1008},2^{1008},2^{1008})$ as our solution to the desired problem.

Note:

Proof for the statement – Square of  odd number equivalent to $1 \pmod{8}$.

Suppose, $n$ is an odd number, so we can write it in the form $n=2k+1$.  Now,         $$n^2\equiv 4k^2+4k+1\equiv 4k(k+1) +1\pmod{8}$$now, observe that $k$ and $k+1$ are two consecutive integers, so one of them is even, hence, we have, $$4k(k+1)\equiv 0\pmod{8} \text{ or, }n^2 \equiv 4k(k+1)+1\equiv 1\pmod{8}.$$

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