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- This topic has 2 replies, 3 voices, and was last updated 2 years ago by Tarit Goswami.

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- July 23, 2018 at 9:35 pm #21702bibhuParticipant
Given,

\( a^2+b^2+c^2+d^2=2^{2018} \)

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

July 28, 2018 at 9:22 pm #21789ARPAN GHOSHParticipantSOLUTIONS 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..

July 30, 2018 at 11:26 pm #21800Tarit GoswamiModeratorWell 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, $$4k(k+1)\equiv 0\pmod{8} \text{ or, }n^2 \equiv 4k(k+1)+1\equiv 1\pmod{8}. $$

- This reply was modified 2 years ago by Tarit Goswami.

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