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

[Albert1]

$p,q,r$ are all primes , and $3p^4-5q^4-4r^2=26$, find $p,q,r$

 

[Opalg]

$p,q,r$ are all primes , and $3p^4-5q^4-4r^2=26$, find $p,q,r$

[sp]Start by reducing mod 5. A square that is not a multiple of $5$ is congruent to $\pm1\pmod5$, and a fourth power is then congruent to $1\pmod5$. So if neither $p$ nor $r$ is $5$, the equation becomes $3 \pm4 = 1\pmod5$, which is impossible. Therefore either $p$ or $r$ must be $5$. If $r=5$ then the equation becomes $3=1\pmod5$, which is still impossible. Therefore $p=5$, so $3p^4= 1875$ and the equation becomes $5q^4 + 4r^2 = 1875 - 26 = 1849.$

For that to be true, $q$ must clearly be odd, so $q\ne2$. The only other prime satisfying $5q^4<1849$ is $q=3$, with $5q^4 = 405$. That leaves us with $4r^2 = 1849-405 = 1444$. So $r^2 = 361$ and thus $r=19.$[/sp]

 

[Albert1]

[sp]Start by reducing mod 5. A square that is not a multiple of $5$ is congruent to $\pm1\pmod5$, and a fourth power is then congruent to $1\pmod5$. So if neither $p$ nor $r$ is $5$, the equation becomes $3 \pm4 = 1\pmod5$, which is impossible. Therefore either $p$ or $r$ must be $5$. If $r=5$ then the equation becomes $3=1\pmod5$, which is still impossible. Therefore $p=5$, so $3p^4= 1875$ and the equation becomes $5q^4 + 4r^2 = 1875 - 26 = 1849.$

For that to be true, $q$ must clearly be odd, so $q\ne2$. The only other prime satisfying $5q^4<1849$ is $q=3$, with $5q^4 = 405$. That leaves us with $4r^2 = 1849-405 = 1444$. So $r^2 = 361$ and thus $r=19.$[/sp]

nice job !

 

[Albert1]

[sp]Start by reducing mod 5. A square that is not a multiple of $5$ is congruent to $\pm1\pmod5$, and a fourth power is then congruent to $1\pmod5$. So if neither $p$ nor $r$ is $5$, the equation becomes $3 \pm4 = 1\pmod5$, which is impossible. Therefore either $p$ or $r$ must be $5$. If $r=5$ then the equation becomes $3=1\pmod5$, which is still impossible. Therefore $p=5$, so $3p^4= 1875$ and the equation becomes $5q^4 + 4r^2 = 1875 - 26 = 1849.$

For that to be true, $q$ must clearly be odd, so $q\ne2$. The only other prime satisfying $5q^4<1849$ is $q=3$, with $5q^4 = 405$. That leaves us with $4r^2 = 1849-405 = 1444$. So $r^2 = 361$ and thus $r=19.$[/sp]

my solution:

Spoiler

as Opalg's mention $p=5$
for $3p^4-5q^4-4r^2=26$
we have:$3p^4-5q^4=4r^2+26>0---(1)$
from (1) we have :$p>q $ and $p,q $ both are odd
$\therefore q=3$
and $\sqrt {3\times5^4-5\times 3^4-26}=\sqrt {1444}=\sqrt {4r^2}=2\times r=38 $
so $r=19$