number theory

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  • #21702
    bibhu
    Participant

    Given,

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

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

    #21789
    ARPAN GHOSH
    Participant

    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
    Tarit Goswami
    Moderator

    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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