I've personally seen this problem everywhere outside of mathlinks, but after searching for it and going through more than 40 pages, I gave up. It's interesting to note that although this is a problem from the Romanian Junior TST from 2001, a weaker version appeared in the 2005 German TST!

This problem is from the third team selection test for the fifth JBMO Buzau, May 20, 2001. I have the book here, if you want I can post the solution.

hvaz wrote: This problem is from the third team selection test for the fifth JBMO Buzau, May 20, 2001. I have the book here, if you want I can post the solution. Yes, I know, I'm posting the problems (edit: and they have been added to the contests section here: http://www.artofproblemsolving.com/Forum/resources.php?c=142&cid=78&year=2001 ). Post the solution if you like!

Taking $\mod 4,8;a,b,c,d$ all even and Infinite descent

It's not infinite descent though is it? $v_2(7\cdot 4^n)$ is finite. I don't want to post another useless post so I'll just post the solution.. If $n=0$ then we obtain the solutions $(2,1,1,1)$ et permutations. If $n\ge 1$ then $a^2+b^2+c^2+d^2=0\pmod{4}$ so $a,b,c,d$ are either all odd or all even. Suppose they are all odd first. Then $7\cdot 4^n=a^2+b^2+c^2+d^2=4\pmod{8}$ so $n\le 1$. We've dealt with $n=0$, for $n=1$ there are the solutions $(5,1,1,1),(4,2,2,2),(3,3,3,1)$ et permutations. Now suppose $a,b,c,d$ are all even, say they are equal to $2a_1,2b_1,2c_1,2d_1$ respectively. Then $a_1^2+b_1^2+c_1^2+d_1^2=7\cdot 4^{n-1}$. Re-iterating this process we see $a_{n-1}^2+b_{n-1}^2+c_{n-1}^2+d_{n-1}^2=28$. We know this has the solutions $(5,1,1,1),(4,2,2,2),(3,3,3,1)$, corresponding to the solutions $(5\cdot 2^{n-1},2^{n-1},2^{n-1},2^{n-1}),(3\cdot 2^{n-1},3\cdot 2^{n-1},3\cdot 2^{n-1},2^{n-1}),(2^{n+1},2^{n},2^{n},2^{n})$ et all permutations.

WakeUp, you haven't considered the case when 2 of a,b,c,d are odd and the other two of them are even

That is impossible since $a^2+b^2+c^2+d^2 \equiv 0 \pmod{4}$.

Pretty late lol but... all number can be written as the sum of four squares right?

Yes all positive integers

Oh wait but according to https://en.wikipedia.org/wiki/Lagrange%27s_four-square_theorem#Uniqueness the numbers in the form of $4 \cdot 7^{n}$ do not belong to the set of number that CANNOT be represented by 4 squares, but how are you gonna prove that tho?

achen29 wrote: Pretty late lol but... all number can be written as the sum of four squares right? correcting myself.. can be written as the sum of NO LESS THAN four squares

For $$n = 0$$we have solutions : $$(2,1,1,1),(1,2,1,1),(1,1,2,1) (1,1,1,2)$$If $$n\geq 1$$then $$a^2+b^2+c^2+d^2 \equiv 0 (mod4)$$, hence $$a,b,c,d$$have the same parity. Now we consider two cases: 1.) If $$a,b,c,d$$are odd numbers,$$ a=2x+1, b=2y+1, c=2z+1, d=2k+1$$Now we have: $$4x(x+1)+4y(y+1)+4z(z+1)+4k(k+1)=4(7.4^{n-1}-1)$$$$n(n+1)$$is a multiple of $$2$$, $$LHS\equiv 0 (mod 8)$$Hence $$7.4^{n-1}-1$$is even. So $$n=1$$: $$a^2+b^2+c^2+d^2=28$$which has the solutions : $$(5,1,1,1),(1,3,3,3)$$and all the permutations... 2.) If $$a,b,c,d$$are even, $$a=2x,b=2y,c=2z,d=2k$$and we get: $$x^2+y^2+z^2+k^2=7*4^{n-1}$$now we proceed recursively. So we obtain the solutions : $$(2^{n+1},2^n,2^n,2^n),(3*2^n,3*2^n,3*2^n,2^n),(2^n,2^n,2^n,5*2^n) $$and all the permutations...