It's USAMO 2015 P5

jelena_ivanchic wrote: It's USAMO 2015 P5 here

My solution: We have $(a^4+b^4)c^2d^2-(c^4+d^4)a^2b^2=(ac+bd)(ac-bd)(ad+bc)(ad-bc)=e^5(cd-ab)(cd+ab).$ We let $ac+bd=p$. We have $3$ cases now. Case 1: $(ac+bd)(ac-bd)(ad+bc)(ad-bc)=e^5(cd-ab)(cd+ab)=0$ This means that $cd=ab$ and either $ac=bd$ or $ad=bc$. If $ac=bd$, we can add the two equations and get $c(a+d)=b(a+d)$, which implies that $c=b$, but this is not true as $a,b,c,$ and $d$ are all distinct. Now if $ad=bc$, we can also add the two equations and we get d(a+c)=b(a+c), which implies that $b=d$, which is also not true. Case 2: $p=e$ We see that $(ac-bd)(ad+bc)(ad-bc)$ has $4$ powers of $e$. Since $ac-bd<p$, $ac-bd$, has $0$ powers of $e$. This means that $ad+bc$ or $ad-bc$ has at least $2$ powers of $e$. Note that $e^2=a^2c^2+2abcd+b^2d^2$. I now claim that $abcd>ad+bc$. Proof: Let $ad$ be $f$ and $bc$ be $g$. Now we can assume that $fg {\le} f+g$. This means that $(fg-g) {\le} f$, so $g(f-1) {\le} f$. We see that if $f>2$, $g=1$, because otherwise, $g(f-1)>f$. If $f=2$, $g=1$ or $2$ will satisfy the equation, but $g>2$ will not. Now since the variables are all distinct, both $f$ and $g$ are not equal to $1$. Therefore, $f=g=2$, but that would mean that two of the variables (a,b,c, and d) are $2$ and the other two are $1$. So this proves our claim that $fg>f+g$. So if $abcd>ad+bc$, both $ad+bc$ and $ad-bc$ are less than $e^2$, so $(ac+bd)(ac-bd)(ad+bc)(ad-bc)$ will not have $5$ powers of $e$ so it will not be equal to $e^5(cd-ab)(cd+ab)$. Therefore, $p{\ne}e$. Now we also see that if $p$ divides $e$, $p^5$ divides $e^5$, so $(ac-bd)(ad+bc)(ad-bc)$ must have $4$ powers of $p$, which as we know doesn't work. So $p$ does not divide $e$. Case 3: $p$ does not divide $e$ and $e^5(cd-ab)(cd+ab)\ne0$ This means that $cd-ab$ or $cd+ab$ are divisible by $p$. Note that $cd+ab{\ge}p$. Now since $p=ac+bd$, we can subtract $cd+ab$ from $ac+bd$ and get $ac-cd+bd-ad=(c-b)(a-d)$. Now this must be less than or equal to $0$. It cannot be equal to $0$, as $a,b,c,d$ are all distinct. If it is less than $0$, then either $c>b$ and $d>a$ or $b>c$ and $a>d$, which both would make it so that $a^4+b^4\ne{c}^4+d^4$, so we have proven that this case doesn’t work. Since all 3 cases don’t work, $ac+bd$ is composite.