Prove, If the sum of the first $n$ prime is also a prime then it is also a hypotenuse of a primitive Pythagorean triples I checked this for all the primitive Pythagorean triples $<300$. 
Some examples would be: 
                                        a. $2+3=5$, 
                                      b. $2+3+5+7=17$, 
                                   c. $2+3+5+7+11+13=41$, 
                             d. $2+3+5+7+11+13+17+19+23+29+31+37=197$
Any kind proof and insight will be appreciated, thanks.
 A: This is false. Euclid’s formula shows that a number is the hypotenuse of a primitive Pythagorean triple if and only if it can be written as the sum of two squares. Fermat proved that an odd prime can be written as the sum of two squares if and only if it’s $1 \pmod 4$. But the sum of the first $60$ primes is $7699$, which is a prime that’s $3 \pmod 4$. (This is the smallest counterexample; see A013918 for a list of sums of primes that are prime.)
A: The standard formula for generating all primitive pythagorean triples is
a=m^2-k^2, b=2km, c=k^2+m^2; c^2=a^2+b^2
where k,m are integers with k in (0,m), gcd(k,m)=1, and either 2 dividing k or 2 dividing m, but not both. Thus, show that your number x which is the sum of the first n primes and prime itself can be written in the form of c, and that there exist numbers(construct them) of the form a and b subject to those constraints. It would be easier to do a couple examples first to get a notion for the flow of the proof. Finally, use induction on n.
