We have $3|3p^4$ and $26 \equiv 2 \pmod{3}$ so $5q^4+4r^2 \equiv 1 \pmod{3}$. We also have $4r^2 \equiv 0,1 \pmod{3}, 5q^4 \equiv 0,2 \pmod{4}$. Therefore $4r^2 \equiv 1 \pmod{3},5q^4 \equiv 0 \pmod{3}$ or $3|q$. We obtain $q=3$. It follows that $3p^4=4r^2+431$. We have $3p^4 \equiv 0,3 \pmod{5}$ and $4r^2+431 \equiv 0,1,2 \pmod{5}$. Thus, $3p^4 \equiv 4r^2+431 \equiv 0 \pmod{5}$ or $5|p$. We get $p=5$. Hence $r=19$. Thus, the answer is $(p,q,r)=(5;3;19)$. $\blacksquare$

Solution: We use modulus $3$. This results in $q^4+2r^2\equiv 2\pmod 3$. It is clear that $q=3$ now. Next, we use modulus $5$. This gives us $3p^4+r^2\equiv 1\pmod 5$. If $p\not = 5$, then $r^2\equiv 3\pmod 5$, contradiction. Hence $p=5$. Now we plug in the knowns and find $r=19$. Our solution is $\boxed{(p,q,r)=(5,3,19)}$. $\blacksquare$

looking mod 3 shows that all of p , q , r can't be >3 A case chase shows that p>3 . the rest is trivial.

nice and easy proof = consider mod $2$ than it is clear that $p,q,r > 2 $ now taking $mod$ $3$ it is clear that $q^4 =0$ $mod$ $3$ thus giving $q=3$ as $q$ is prime now taking $mod$ $5$ shows that and considering $p\not = 5$ gives $r^2 =3$ $mod$ $5$ a contradiction ! hence $p = 5 $ and thus $r=19$ thus $(p,q,r) = (5,3,19) $

Nice and easy solution

By mod3, get q=3, so we are left with 3p^4-4r^2=431 Substituting, neither of p=2 and r=2 gives a solution Therefore, p and r are odd Therefore, p^4 is 1 or 5 mod 10 so 3p^4 is 3 or 5 mod 10 Also, r^2 is 1 or 5 or 9 mod 10, so 4r^2 is 4 or 0 or 6 mod 10 Trying all 6 cases, we get that only 3p^4 congruent 5mod10 and 4r^2 congruent 4mod10 make their difference 1mod10 Therefore, p^4 is 5mod 10 Therefore, p=5 then solving, get r=19 So only (p,q,r)=(5,3,19) Is this right?

Practicing for when I join JBMO this June

Taking $\pmod{5}$ and using Fermat's Little Theorem , we have $$3p^{4}-5q^{4}-4r^{2} \equiv \{3,0\} - \{0,4,-4\} \equiv 1 \pmod{5}$$The only combination which satisfies the condition is $$(p^4 ,r^2) \equiv (0,4) \pmod{5} \quad \text{since} \quad 0 - 4 \equiv -4 \equiv 1 \pmod{5}$$$$\implies 5 \mid p^4 \implies 5 \mid p \implies 5 = p$$Putting $p=5$ in our original equation , we have $$5q^4 + 4r^2 = 1849 = 43^2$$$$\implies 5q^4 = (43+2r)(43-2r)$$$$\gcd(43+2r , 43 - 2r) = \gcd(43+2r , 86) = 1,2,43,86$$But $\gcd$ cannot be $2,86$ since $43 +2r = \text{odd}$ and it cannot be $43$ either because if $\gcd = 43 \implies r = 43$ but $4\cdot 43^2 >> 1849 \implies \gcd = 1 \implies 5$ will go into 1 factor and $q^4$ in the other factor Now $q^4 \ge 16 > 5$ and $43 + 2r > 43 - 2r$ Thus 5 will go into the smaller factor , i.e $$5 = 43 - 2r$$$$\implies 2r = 38 \implies r = 19$$$$\implies q^4 = 43 + 2r = 43 + 38 = 81 \implies q = 3$$$$\boxed{(p,q,r) = (5,3,19)} \quad \blacksquare$$

For storage. Classic modulo application. Claim. $p=5$ Modulo $5$ takes us to congruence $3p^{4}+r^{2}\equiv 1\pmod 5$. Suppose $p\neq 5$, then by Fermat's little theorem, $3p^{4}\equiv 3\pmod 5$, which means that $r^{2}\equiv 3\pmod 5$, but that's a contradiction, since $r^{2}\equiv -1,0,1\pmod 5$ for any $r$. Therefore, $p=5$. Claim. $q=3$ Modulo $3$ takes us to congruence $q^{4}+2r^{2}\equiv 2\pmod 3$. Suppose $q\neq 3$, then by Fermat's little theorem, $q^{4}\equiv 1\pmod 3$, which means that $r^{2}\equiv 2\pmod 3$, but that's a contradiction, since $r^{2}\equiv 0,1\pmod 3$ for any $r$. Therefore, $q=3$. Now we solve equation for $r$ and obtain $r=19$. Answer. $\boxed{(p,q,r)=(5,3,19)}$. The motivation of claims? Well, since it is given that $p,q,r$ are primes and we are given squares of primes and coefficients of those like $3,4,5$, we recognize modules $3,4,5$ and we obviously expect to have some "easy" solution triplet or obtain no solutions at all, therefore we also might expect solutions like $2,3,5,7$. Eventually, examing modules $3$ ja $5$, turns out that those solve the problem with some easy "case" anaylsis.

The equation$\pmod3$ is $q^4\equiv2+r^2$, and since $q^4,r^2\equiv0,1$ we need to have $3\mid q^4$, so $q=3$. Then we have $3p^4-4r^2=431$. By$\pmod5$, $3p^4+r^2\equiv1$, and since $p^4,r^2\equiv0,1,4$ we need $p=5$. This gives the unique solution $(p,q,r)=\boxed{(5,3,19)}$.

I used mod $5$ since $5$ has only 2 possible modulos for $x^4$.So 2 cases arise and we are done .PLUS: working withmodulo 5 was more fun than modulo 3.

first taking mod 3 we get 5q^4+4r^2 = 1 (mod 3) <=> q = 0 (mod 3) <=> q=3 now plugin in that q=3 <=> 3p^4-4r^2= 431 and taking this (mod 5) => 3p^4-4r^2=1 (mod 5), since only 0,1 and 4 are quadratic remainders (mod 5) only solution is p = 0 (mod 5) so p=5 now we have q=3 and p=5 <=> r=19; so only solution is (5,3,19)

Mod 3 and mod 5

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Taking modulo 3, we see that $$3p^4-5q^4-4r^2 \equiv q^4-r^2 \equiv -1 \pmod 3.$$Since squares can only be $0$, $1$ modulo 3, we see that $q^4 \equiv 0 \pmod 3$, $r^2 \equiv 1 \pmod 3$; but since $p$, $q$, $r$ are primes, $q=3$. Thus, the equation simplifies to $3p^4-4r^2=431$. Taking modulo 5, we have $$3p^4-4r^2 \equiv 3p^4+r^2 \equiv 1 \pmod 5.$$If $p \not\equiv 0 \pmod 5$, by Fermat's Little Theorem, we get that $$3+r^2 \equiv 1 \pmod 5 \implies r^2 \equiv 3 \pmod 5,$$which has no solutions. Then, if $p \equiv 0 \pmod 5$, we get that $$r^2 \equiv 1 \pmod 5 \implies r \equiv 1, 4 \pmod 5,$$meaning that $p=5$. Plugging it back in to the original equation, we get $r=19$, so the only triple that satisfies the equation is $(5, 3, 19)$.

$x\equiv 0,1.2,3,4 \pmod{5} \implies x^2\equiv 0,1,-1 \pmod{5} \implies x^4\equiv 0,1 \pmod{5}$ $26\equiv 1, \pmod{5}$ $4r^2=\equiv 0,1,-1 \pmod{5}$ $5q^4\equiv 0,1,-1 \pmod{5}$ $\implies$ $ 26+4r^2+5q^4\equiv 0,3,4 \pmod{5} \implies $ $3p^4\equiv 0,3,4 \pmod{5}$ $...(*)$ Earlier we found that $x^4\equiv 0,1 \pmod{5} \implies 3x^4\equiv 0,1 \pmod{5} \implies 3p^4\equiv 0,1 \pmod{5}$ $...(**)$ Combining $(*)$ with $(**)$ we get that $3p^4\equiv 0 \pmod{5} \implies 5 \mid 3p^4 \implies 5 \mid p^4 \implies 5 \mid p$ $\implies$ $p=5$ $3p^{4}-5q^{4}-4r^{2}=26 \implies 5q^4+4r^2=1849 \implies 5q^4+4r^2=43^2 $ If $q \ge 7$ then $5q^4+4r^2 > 43^2$ So $q < 7$ we check the cases and we find that only $q=3$ and $r=19$ works So $(p,q,r)=(5,3,19)$

For mod 3; 5q⁴+4r² = -2 (mod 3) 5q⁴+4r² = 1 (mod 3) 2q⁴+r²=1(mod 3) So, 3/q⁴ let's assume that q=3. And r²=1 (mod 3) For mod 5; 3p⁴-4r²= 1 (mod 5) r= 1,-1 (mod 5) First case; i) r= 1( mod 5) We have solution 3p⁴-4=1 (mod 5) 3p⁴=0 (mod 5) 5/p So, p=5 and r=19 ii) r= -1 ( mod 5) there's no solution. The only solution is (p,q,r) ===> (5,3,19)

firs take mod 3 rest is easy