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

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SUMMARY

The equation \(3p^4 - 5q^4 - 4r^2 = 26\) requires finding prime numbers \(p\), \(q\), and \(r\). Through analysis, the only valid solution is \((p, q, r) = (2, 1, 1)\), where \(p\) and \(q\) are the primes 2 and 1, respectively, and \(r\) is also 1. This solution satisfies the equation, confirming that the primes must be carefully selected to meet the criteria of the equation.

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Albert1
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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 : $q^4-r^2\equiv -1 \,\,(mod \,\, 3)$
 

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