MHB Solving for $(p,q,r)$ When $3p^4-5q^4-4r^2=26$

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The equation 3p^4 - 5q^4 - 4r^2 = 26 requires finding prime numbers p, q, and r that satisfy it. The discussion revolves around testing various combinations of prime numbers to solve for the variables. Participants suggest methods for narrowing down potential values, emphasizing the need for primes in the solution. The challenge lies in balancing the equation while adhering to the constraints of primality. Ultimately, the goal is to identify the specific triplet (p, q, r) that fulfills the equation.
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$p,q,r$ are all primes ,and $3p^4-5q^4-4r^2=26$

find $(p,q,r)=?$
 
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Albert said:
$p,q,r$ are all primes ,and $3p^4-5q^4-4r^2=26---(1)$

find $(p,q,r)=?$
hint:
(1) mod 3 we have : $q^4-r^2\equiv -1 \,\,(mod \,\, 3)$
 

Thread 'Area of Overlapping Squares'

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