Day 15: Squaring the primes

Because the sum is even, one of the primes must be 2. (Not all three can be 2).

WLOG assume that p=2. Then $q^2+r^2=2010$ . All squares are 0 or 1 mod 3. 2010 is 0 mod 3, so both squares must also be 0 mod 3. However, this means that both primes must be 3, which does not work. Therefore, no solutions exist.

First note that $2$ is not a possible value of $p$ . If it were, $2010$ would be expressible as a sum of two squares. However, only numbers of the form $2^k \cdot a \cdot b^2$ , where $a$ is a product of primes congruent to $1$ mod $4$ and $b$ is a product of primes congruent to $3$ mod $4$ , can be represented as a sum of two squares, and $2010$ does not have this form. Thus, $2$ is not a possible value of $p$ .

Therefore, $p$ , $q$ , and $r$ are all odd, so $p^2+q^2+r^2$ is odd. However, $2014$ is even. Thus, no such primes exist, so the sum is $0$

All primes are $1 \bmod 3$ except for $3$ which is $0 \bmod 3$ . Now notice that the RHS is $1 \bmod 3$ . Therefore the primes' squares must be $1, 0, 0 \bmod 3$ in some order. But that means that $p^2 = 1996$ (for some prime $p$ ) for which there are no solutions.

Therefore there are no solutions and so the answer is $\boxed{0}$ .

every prime no. can be written in the form of either 6n+1 or 6n-1 ; now if we see that whole squaring either of the squares, we get 1 as the remainder but 2014 is exactly divisible by 2014 so none of the values of "p", "q",or "r" are possible. note-: the rule that prime no's are in the form 6n=1 or 6n-1 is applicable for all primes greater then or equal to 5.

First of all,we know that all the primes, $p,q,r$ can't be odd as if they are,then $L.H.S$ would be odd $\Rightarrow$ one of the primes has to be $2.$ Let us say that $p=2.\Rightarrow\ q^2+r^2=2010.$ Now,all squares are $\equiv0,1,4,9\pmod{16}.$ In this case $L.H.S$ is $\equiv10\pmod{16}.$ The only combination that fits is $1,9.$ Let us say that $q\equiv1\pmod{16}.$ Now,for $r$ we have these possibilities, $\begin{cases}\\16y+3\\16y+5\\16y+11\\16y+13\end{cases}$ Checking the first case gives, $256(x^2+y^2)+32x+96y=2000.$ Dividing both sides by 16 gives, $16(x^2+y^2)+2x+6y=125.$ Now,the $L.H.S$ is divisible by $2$ but the $R.H.S$ isn't.Thus,no solutions exist.A similar thing would happen for all the other cases.Thus,no solutions exist.