This is a prime problem

Using Fermat's square prime Theorem: If ${p}$ = $m^{2}$ + $n^{2}$ , then ${p}$ $\equiv1$ mod(4). Searching for primes is relatively simple, so we do 5+13+17+29+37+41+53+61+73+89+97 = 515 $\equiv22$ mod(29). (Note: 2 is an exception to the rule, but m=n is forbidden in the question)

We can easily find than $p=2$ is not a solution. Now, Fermat's theorem on sums of two squares tells that an odd prime $p$ can be expressed as a sum of two squares if and only if $p \equiv 1 \pmod{4}$ . The squares must be distinct because one of them must be even and one of them must be odd.

Hence, the answer is $5+13+17+29+37+41+53+61+73+89+97 \equiv 515 \equiv \boxed{22} \pmod{29}$ .

Squares are:
$m^2$ :- 1, 9, 25, 49, 81 and
$n^2$ :- 4, 16, 36, 64.

Sum of 'p's is : 5+13+17+29+37+41+ 53+61+73+89+97 =515....515 (mod 29) =22.
$\boxed { 22 }$
I am adding this now on 11/6/2014.

We can find 5,13,17,......97 through a table. Column with $m^2$ when ' m 'as odd (1, 3, 5, 7, 9), and row with $n^2$ when 'n' as even (2, 4, 6, 8) since p< 101.
Put the sum of column and row at an intersection if it is a prime or X if it is not a prime or it is greater than 100. We will have X at nine places (25, 45, 65, 85, 85, 114, 117, 145). Remaining numbers as given above. Add them up and (mod 29) = 22.
There is a little long method but intresting.
While filling up the table, for every item in the first column and row,have the
(mod 29) of it in the paranthises. At the intersection also have a parenthises that contain the sum of column paranthises and row parenthises. Finaly add only the numbers in the parenthises algebrically. The sum (mod 29) will be = 22.
For example intersection of column with 25 and row 16 would look like
.............................25(-4)
16(16)............... 41(12)