Take the equation 3p^4-5q^4-4r^2=26 mod 3. Also let q'=q^4 mod 3 and r'=r^2 mod 3; q' and r' each can be either 0 or 1.
Then the equation becomes q' + 2r' = 2 mod 3. Of the four possible (q',r') pairs only (0,1) satisfies the modular equation.
q is then a multiple of 3, but q is given to be a prime; therefore
q=3.
Same thing, this time mod 5. Let p"=p^4 mod 5 and r"=r^2 mod 5; p" can be either 0 or 1 and r" can be either 0, 1, or 4.
Then the equation becomes 3p" + r" = 1. Of the four possible (p",r") pairs only (0,1) satisfies the modular equation.
p is then a multiple of 5, but p is given to be a prime; therefore
p=5.
With two values known, make a direct substitution: 3*5^4 - 5*3^4 - 4r^2 = 26. Then 4r^2 = 1444 makes r=19, which is prime. Thus the unique prime triplet satisfying the equation is (p,q,r) = (5,3,19).
Solution
