[Tarit Goswami]

We proceed using induction on the degree of  \(p\). It's evident that  \(p\) must have even degree, and the proposition is trivial for \(p\)  constant. Since  \(p\) has even degree, it has a non-negative global minimum, say \(m\) . Then \( p-(\sqrt{m})^2 \)  is tangent to the x-axis and so is divisible by \((x-c)^2\)  for some real \(c\) . Then \(\frac{p-(\sqrt{m})^2}{(x-c)^2} \)  is expressible as a sum of squares by assumption, and we simply need to multiply through by  \( (x-c)^2 \) and add  \(\sqrt{m}^2\) to get that \(c\)  is as well.

This argument can be easily refined to again get that  \(k=2\) suffices.

[Tarit Goswami]

Yes, Shantanu is correct, they want to say,

pairwise distinct and no common factor

i.e; it is mendetory to consider that they are paurwise co-prime. Otherwise, it is pointless to say that the all seven have "no common factor", when they have said that they are "pairwise distinct". I will come back with solution. Give it a try again.

[Tarit Goswami]

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, $$4k(k+1)\equiv 0\pmod{8} \text{ or, }n^2  \equiv 4k(k+1)+1\equiv 1\pmod{8}. $$

[Tarit Goswami]

First observe that length of the segment \( AH = 2R\cos{\angle{BAC}} \) , where \( R\) is the circumradius.

So, $$AH = 2\cdot \cos{60^o} = 1$$

[Tarit Goswami]

Check this out(the last solution). Let me know if face any problem to understand.

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