# Doubt from number theory

#1

Find the smallest number n such that there exist polynomials f_1, f_2, \ldots , f_n with rational coefficients satisfying [x^2+7 = f_1\left(x\right)^2 + f_2\left(x\right)^2 + \ldots + f_n\left(x\right)^2.]

#2

hari shankar sirr please see this problem..

#3

#4

Is answer 5 ? , if yes I will share solution soon , but first tell me the source . and btw this question is above IMO lvl I think

#5

its imo level question my sir asked me..

#6

Gud , but first target for INMO Sneha , do your best for INMO , camp is sufficient for IMO lvl and also professors will help u. ( and is my answer correct ?)

#7

ya..

#8

Will share solution soon, btw u tried it ??

#9

ya i tried it...

#10

#12

will share latter

#13

x²+ 2² +1²+1²+1² therefore n>=5

#14

it may be 1 approch i thought that but their may be proper method for this..

#15

Use the diophantine equation a²+b² +c² =7d²

#16

right..

#17

Use infinite descent on that Diophantine equation , easily proves for n= 4 (non achievable case)

#18

For a standard proof without using infinite descent : Assume n=4 is possible, then there exist integers a,b,c,d,e,f,g,h,m (whose gcd is 1) such that (ax+b)^2+(cx+d)^2+(ex+f)^2+(gx+h)^2=m^2(x^2+7)

Take x=1,3,5, we get (a+b)^2+(c+d)^2+(e+f)^2+(g+h)^2=8m^2\quad...(1) (3a+b)^2+(3c+d)^2+(3e+f)^2+(3g+h)^2=16m^2\quad...(2) (5a+b)^2+(5c+d)^2+(5e+f)^2+(5g+h)^2=32m^2\quad...(3) Now we use the following lemma: If 8|p^2+q^2+r^2+s^2, then p,q,r,s are even. The proof is trivial.

Using the lemma on (3) twice, we get that 4|5a+b,5c+d,5e+f,5g+h. Hence 4|a+b,c+d,e+f,g+h. Combining with (1), we get that m is even. Using the lemma on (2) twice, we get that 4|3a+b,3c+d,3e+f,3g+h. So a,c,e,g are even, which implies that a,b,c,d,e,f,g,h,m are all even, contradicting the assumption that their gcd is 1 .

1 Like
#19

#20

Actual me , little bit. U have any other method Samrat ?

#21

General case ( without assuming one to be x^2 only.) (Just the lemma that 7 can't be expressed as a sum of three squares) : (using vectors ) assume it's possible for n=4.

Then, letting the coefficients be a_i,b_i as above, we have

a_1^2+a_2^2+a_3^2+a_4^2=1
b_1^2+b_2^2+b_3^2+b_4^2=7
a_1b_1+a_2b_2+a_3b_3+a_4b_4=0

Consider the two 4-dimensional vectors \overrightarrow{u}=(a_1,a_2,a_3,a_4) and \overrightarrow{v}=(b_1,b_2,b_3,b_4)

Then, we have that |u|=1, |v|=7, and that the two are orthogonal.

Consider the transformation represented by a matrix which takes \overrightarrow{u} to \overrightarrow{x}=(1,0,0,0). This transform preserves norm and orthogonality, and, as the matrix necessarily has rational entries, due to its entries being solutions to a system of equations with rational coefficients, preserves rationality (the exact form of the matrix isn't too hard to find either).

This will then take \overrightarrow{v} to \overrightarrow{y}=(0,c_2,c_3,c_4), with |y|=7. But, no 3 rational numbers have squares which add to 7, so no such v' can exist, which implies no such u,v can exist, so 4 is impossible