Olympiad stuff

suppose we take 49 and we write it as
49=1⁴+1⁴+....so on 49 times then this proves to be wrong.....

i think the question wants us to prove that we require maximum only 48 integers each raised to power 4 , to express any other positive integer ..........is it correct ??

btw i dont understand why dey use very high language in questions .....
questions are asked to test a student's intellectual capability or their language skills ?? i m serious about it ..i hav gone thru many physics problems which will take 2 days to understand the grammar of d question and the answer requires no more than 2 seconds of thinking ...... LOL ...

qwerty is correct. The solution is ----

(a+b)⁴ + (a-b)⁴ = a⁴ + 4a³b + 6a²b² + 4ab³ + b⁴ + a⁴ - 4a³b + 6a²b² - 4ab³ + b⁴
= 2a⁴ + 12a²b² + b⁴

so, [(a+b)⁴ + (a-b)⁴] + [(a+c)⁴ + (a-c)⁴] + [(a+d)⁴ + (a-d)⁴] + [(b+c)⁴ + (b-c)⁴] + [(b+d)⁴ + (b-d)⁴] + [(c+d)⁴ + (c-d)⁴] = 6a⁴ + 6b⁴ + 6c⁴ + 6d⁴ + 12a²b² + 12a²c² + 12 a²d² + 12b²c² + 12b²d² + 12c²d²
= 6 ( a² + b² + c² + d² )²

If you carefully see , then you will find that the l.h.s of the identity contains 12 integers in the form of fourth powers .
But each one of them can again be expressed in terms of different a,b,c,d in the form 6 x (the sum of squares of different a,b,c,d)².
So each term in the l.h.s can be expressed with maximum of different 4 numbers each raised to the power 4.
[as (a² + b² +c² +d²)² contains 4 terms with powers 4.]
So r.h .s = 12 x maximum 4 different integers each raised to the power 4
or the initial number = 48 integers each raised to the power 4