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 'Erroneously finding discrepancy in transpose rule'

Thread 'Erroneously finding discrepancy in transpose rule'

Obviously, there is something elementary I am missing here. To form the transpose of a matrix, one exchanges rows and columns, so the transpose of a scalar, considered as (or isomorphic to) a one-entry matrix, should stay the same, including if the scalar is a complex number. On the other hand, in the isomorphism between the complex plane and the real plane, a complex number a+bi corresponds to a matrix in the real plane; taking the transpose we get which then corresponds to a-bi...
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