Find all triples $(k, m, n)$ of positive integers such that $m$ is a prime and: (1) $kn$ is a perfect square; (2) $\frac{k(k-1)}{2}+n$ is a fourth power of a prime; (3) $k-m^2=p$ where $p$ is a prime; (4) $\frac{n+2}{m^2}=p^4$.
Problem
Source: Serbia JBMO TST 2023 P4
Tags: number theory
grupyorum
29.05.2023 20:34
I claim there are no solutions. Assume the contrary and see that $k=m^2+p$ and $n = m^2p^4-2$. We first show $m$ is odd and $p\equiv 3\pmod{8}$. To that end, assume first $m=p$. Then $kn = p(p+1)(p^6-2)$ so that $kn$ being a square implies $p\mid p^6-2$, that is $p=2$. Clearly $m=p=2$ does not fulfill the condition, so $m\ne p$. Now if one of $m,p$ is even, then $m^2+p$ is odd whereas $m^2p^4-2\equiv 2\pmod{4}$, so $v_2(kn)=1$, absurd. Hence $m\ne p$ and $m,p$ are both odd. Next, letting $kn = a^2$ for some $a$, we have $a^2\equiv -2m^2\pmod{p}$, and thus $(-2/p)=1$ (as $p\nmid m$). So, $p\in\{1,3\}\pmod{8}$. If $p\equiv 1\pmod{8}$ then $k=m^2+p\equiv 2\pmod{8}$ whereas $n=m^2p^4-2\equiv -1\pmod{8}$, so $kn$ is not a square. Thus, $p\equiv 3\pmod{8}$.
Now, we use the condition $k(k-1)/2 + n$ is a fourth power of a prime. Clearly this prime is not 2 by size considerations, so $k(k-1)/2 + n\equiv 1\pmod{8}$. Now, as $k=m^2+p\equiv 4\pmod{8}$, we set $k=8i+4$ for some $i$. Then, $k(k-1)/2 = 32i^2+28i+6$. So, $k(k-1)/2+ n \equiv 4i+1\equiv 1\pmod{8}$. In particular, $8\mid 4i$ so that $i$ is even. Lastly, set $i=2t$ to get $k = 16t+4$. Thus, $kn = 4(4t+1)n$, where $n=m^2p^4-2\equiv -1\pmod{4}$. This however is a contradiction.
Remark. The second half of my proof (redacted above) is wrong.
a_507_bc
29.05.2023 20:48
$(28, 5, 2023)$ works?
gvole
29.05.2023 22:48
Olympiad math doesn't get much better than the TST's in Serbia this year...
P2nisic
30.05.2023 01:32
a_507_bc wrote: Find all triples $(k, m, n)$ of positive integers such that $m$ is a prime and: (1) $kn$ is a perfect square; (2) $\frac{k(k-1)}{2}+n=r^4$ is a fourth power of a prime; (3) $k-m^2=p$ where $p$ is a prime; (4) $\frac{n+2}{m^2}=p^4$. Let $d=(n,k)$. If $2|d$ then from (3),(4) we get $p=2$ so $k=m^2+2,n=16m^2-2$ but then: $(2m+1)^2<kn<(2m+2)^2$ for every $m>1$ for $m=1$ we get no sollution. If $d=odd$ then by (2) we have $d=1,r,^2,r^3,r^4$ If$d=r^3,r^4$ then $k(k-1)>2r^4$ contradiction on (2) If $d=1,r^2$ using (1) we have $n=a^2$ but then at (4) we have $a^2+2=b^2$ contradiction So $d=r$ with (1) gives $n=a^2r,k=b^2r$ rewrite the condition as: (5)$a^2r+2=p^4(b^2r-p)$ with $b^2r-p=m^2$ (6) $b^4r+2a^2-b^2=2r^3$ If $3|a$ we get contradition at (5). So now at (6) by considering to cases for $b$ we get in both that $r=1(mod3)$ Going back at (5) gives $3|LHS$ so $3|p^4(b^2r-p)$ If $p=3$ we get the sollution $(k,m,n)=(28,5,2023)$(see that $r|245$). Otherwise let $3|b^2r-p$ we get $p=1(mod3)$ Consider now the cases $a^2=1,4,7(mod9)$ all gives contradiction. For $a^2=1(mod9)$ we have: $3|m$ so $9|a^2r+2$ which means $r=7(mod9)$ then back on (6) we get:$7b^4+2-b^2\equiv 2*7^3=2(mod 9)\Leftrightarrow 7b^4-b^2\equiv 0(mod9)\Leftrightarrow b^2\equiv 4(mod9)$ But $p\equiv -b^2r\equiv -4*7\equiv 8(mod9)$ contradiction since $p=1(mod3)$ Similar for $a^2=4,7(mod9)$