Either \frac{n^4+m}{m^2+n^2} or \frac{n^4-m}{m^2-n^2} is integer

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Either \frac{n^4+m}{m^2+n^2} or \frac{n^4-m}{m^2-n^2} is integer

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Source: 2023 Turkey Junior National Olympiad P3

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#1 h Dec 22, 2023, 8:21 AM

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Let $m,n$ be relatively prime positive integers. Prove that the numbers
$\frac{n^4+m}{m^2+n^2} \qquad \frac{n^4-m}{m^2-n^2}$ cannot be integer at the same time.

This post has been edited 1 time. Last edited by AlperenINAN, Dec 22, 2023, 8:24 AM

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#4 h Dec 22, 2023, 12:01 PM • 1 Y

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Suppose the contrary. Note that $n^2\equiv -m^2\pmod{m^2+n^2}$ , so that $n^4+m\equiv n^2\cdot n^2+m\equiv -m^2n^2+m\pmod{m^2+n^2}$ . So, $m^2+n^2\mid m(mn^2-1)$ . As $(m,m^2+n^2)=1$ , we thus obtain $m^2+n^2\mid mn^2-1$ . Likewise, $m^2-n^2\mid mn^2-1$ .Now, observe that $(m^2+n^2,m^2-n^2)\in\{1,2\}$ . So,
$m^4-n^4\mid 2(mn^2-1)\Rightarrow |m-n|\cdot (m+n)(m^2+n^2)\le 2mn^2-2.$ Simple inequalities show that in fact
$(m+n)(m^2+n^2)+2>2mn^2,$ yielding a contradiction.

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#5 h Dec 22, 2023, 11:05 PM

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My solution is same but in a longer way:

Using the equality $n^4 + m = n^4 - m^4 + m^4 + m = (n^2-m^2)(n^2 + m^2) + m(m^3 + 1)$ , we have $m^2 + n^2 \mid n^4 + m \Rightarrow m^2 + n^2 \mid m(m^3 + 1) \qquad (1)$ Since no positive divisor of $m$ divides $m^2 + n^2$ , $m$ and $m^2+n^2$ are coprime. From $(1)$ , we obtain $m^2 + n^2 \mid m^3 +1 \qquad (2)$ .
Using the equality $n^4 - m = n^4 - m^4 + m^4 - m = (n^2-m^2)(n^2 + m^2) + m(m^3 - 1)$ , we have $m^2 - n^2 \mid n^4 - m \Rightarrow m^2 - n^2 \mid m(m^3 - 1) \qquad (3)$ .
Since no positive divisor of $m$ divides $m^2 - n^2$ , $m$ and $m^2 - n^2$ are coprime. From $(3)$ , we obtain $m^2 - n^2 \mid m^3 - 1 \qquad (4)$ .

Similarly, since $m^3 - 1 = m(m^2 - n^2) + mn^2 - 1$ , we have $m^2 - n^2 \mid mn^2 - 1$ and since $m^3 + 1 = m(m^2 + n^2) - mn^2 + 1 = m(m^2 + n^2) - (mn^2 - 1)$ , we have $m^2 + n^2 \mid mn^2 - 1$ .

The greatest common divisor of $m^2 - n^2$ and $m^2 + n^2$ is $1$ or $2$ .
To see this, let's write $m^2 + n^2 = dk_1$ and $m^2 - n^2 = dk_2$ and solve the system: $2m^2 = d(k_1 + k_2)$ and $2n^2 = d(k_1 - k_2)$ .
If $d>2$ , then $d \mid 2m^2$ and $d \mid 2n^2$ , which contradicts the coprimality of $m$ and $n$ .

Since $m^2 + n^2 \mid mn^2 - 1$ and $m^2 - n^2 \mid mn^2 - 1$ , if $\text{gcd} (m^2 + n^2, m^2 - n^2) = 1$ , then $(m^2 + n^2)(|m^2 - n^2|) \mid (mn^2 - 1)$ ,
and if $\text{gcd} (m^2 + n^2, m^2 - n^2) = 2$ , then $(m^2 + n^2)(|m^2 - n^2|) \mid 2(mn^2 - 1)$ .

In either case, we have $|m^4 - n^4| \leq 2(mn^2 - 1)$ .

For $m > n$ , we have $m^4 - (m-1)^4 \leq m^4 - n^4 \leq 2(mn^2 - 1) \leq 2(m^3 - 1)$ .
For $n > m$ , we have $n^4 - (n-1)^4 \leq n^4 - m^4 \leq 2(mn^2 - 1) \leq 2(n^3 - 1)$ .
(Note that for $m^2 - n^2$ to be defined, $m^2 - n^2 \neq 0 \Longrightarrow m \neq n$ , so the case $m=n$ is excluded.)

Both inequalities are essentially the same.
If we solve for $k = m$ or $k = n$ , we get $\begin{array}{rcl} k^4 - (k-1)^4 \leq 2k^3 - 2 &\Longrightarrow& 4k^3 - 6k^2 + 4k - 1 \leq 2k^3 -2 \\ &\Longrightarrow & 2k^3 < 2k^3 + 4k + 1 \leq 6k^2 \\ &\Longrightarrow & k < 3 \end{array}$ which implies that the larger of $m$ and $n$ must be less than $3$ . Therefore, we only need to check the pairs $(m,n)=(2,1)$ or $(m,n) = (1,2)$ .
It can be easily seen that these do not satisfy the condition. Thus, in this case, the numbers cannot both be integers.

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#6 h Dec 23, 2023, 12:18 AM

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combo incompetency

Notice that
$m^2+n^2\mid n^4+m\implies m^2+n^2\mid mn^4+m^2\implies m^2+n^2\mid mn^4-n^2\implies m^2+n^2\mid mn^2-1$ $m^2-n^2\mid n^4-m\implies m^2-n^2\mid mn^4-m^2\implies m^2-n^2\mid mn^4-n^2\implies m^2-n^2\mid mn^2-1$ now since
$\gcd{(m^2+n^2,m^2-n^2)}=\gcd{(m^2+n^2,2n^2)}$ we can note: if $m^2+n^2$ is odd then this is $1$ , otherwise it's $2$ . Importantly,
$k\cdot \text{lcm}(m^2+n^2,m^2-n^2)=|m^4-n^4|$ where $k\mid 2$ hence
$m^4-n^4\mid 2mn^2-2.$ (Rip bad wording lmao) Anyway, if $n>m$ then clearly
$n^4-m^4\le 2mn^2-2\implies n^4-(n-1)^4\le 2(n-1)n^2-2\implies 4n^3-6n^2+4n-1\le 2n^3-2n^2-2$ $2n^3-4n^2+4n+1\le 0$ $2n^3-4n^2+4n<0$ $n^2-2n+2<0$ $(n-1)^2+1<0$ which obviously doesn't work. If $n=m$ then we're dividing by $0$ in the original problem, which doesn't work. If $n<m$ then
$m^4-(m-1)^4\le 2m(m-1)^2-2\implies 4m^3-6m^2+4m-1\le 2m^3-4m^2+2m-2$ $2m^3-2m^2+2m+1\le 0$ $2m^3-2m^2+2m<0$ $m^2-m+1<0$ $m(m-1)+1<0$ again we fail. Done.

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Primeniyazidayi

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Primeniyazidayi

#7 h Apr 11, 2025, 11:24 AM

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My solution in the contest:
FTSOC we assume that both of them are integer.Then $\frac{n^4+m}{m^2+n^2}+\frac{n^4-m}{m^2-n^2}=\frac{2m^2n^4-2mn^2}{m^4-n^4}$ is also an integer.Because $(m,n)=1$ ,we have that $(mn^2,m^4-n^4)=1,$ so $\frac{2mn^2-2}{m^4-n^4}$ is also an integer.Note that $2mn^2-2 > 0.$ Then we must have that $2mn^2-2 \geq \left| m^4-n^4 \right|$ If m>n,then for a fixed n we should have $f(m)=m^4-2mn^2-(n^4-2) \leq 0$ But $f'(m)=4m^3-2n^2>0$
and $f(n+1)=n^4+4n^3+6n^2+4n+1-2n^3-2n^2-n^4+2>0$ ,contradiction.If n>m,then for a fixed m we must have $f(n)=n^4-2mn^2-m^4+2 \leq 0$ .But we have that $f'(n)=4n^3-4mn>0$ and $f(m+1)=m^4+4m^3+6m^2+4m+1-2m^3-4m^2-2m-m^4+2>0,$ a contradiction once again.Thus we are done.

This post has been edited 1 time. Last edited by Primeniyazidayi, Apr 11, 2025, 11:25 AM

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