Oops I fake solved it hoping to get 1 pt Take mod 4 and do casework and then for one case, "switch c and d" so that ac+bd is even. I know it is not allowed but am hoping to get 1

stephcurry wrote: Oops I fake solved it hoping to get 1 pt Take mod 4 and do casework and then for one case, "switch c and d" so that ac+bd is even. I know it is not allowed but am hoping to get 1 Exactly what I did, except mod 16... Edit: Apparently it's "well known": http://en.wikipedia.org/wiki/Beal%27s_conjecture "The case (x, y, z) = (2, 4, n) was proved for n ≥ 4 by Michael Bennet, Jordan Ellenberg, and Nathan Ng in 2009."

stephcurry wrote: Oops I fake solved it hoping to get 1 pt Take mod 4 and do casework and then for one case, "switch c and d" so that ac+bd is even. I know it is not allowed but am hoping to get 1 No, I doubt this is going to get you any points. Does anyone have a solution to this? I tried for a long while today and got nowhere.

Oh wow if you didn't solve it then maybe nobody did (or very few )

The only real progress I made it seems, is taking their product and using the well known trick with products of sums or squares, which is pretty trivial observation; what leads on from this is bashing/factoring/casework assuming $(ac+bd)$ is prime, but it didn't get anywhere... $(ac+bd)^{2}(ac-bd)^{2} + (a^{2}d^{2}+b^{2}c^{2})^{2} = e^{10}$

AkshajK says you can solve it by factoring or something idk apparently it factors nicely

One approach is to use the fact that $ac+bd$ divides $$a^4c^4-b^4d^4=(e^5-b^4)c^4-b^4(e^5-c^4)=e^5(c^4-b^4).$$Now we get that if $ac+bd$ is prime, then $ac+bd$ must divide at least one of $e, b+c, b-c, b^2+c^2.$ The first three cannot be true by size, and for the next, assuming for contradiction, we get that $ac+bd$ divides $$a(b^2+c^2)-c(ac+bd)=b(ab-cd).$$Since $ac+bd$ does not divide $b$ then we have that $ac+bd$ divides $ab-cd.$ Without loss of generality assume $ab>cd$ so that $ac+bd\le ab-cd$ or $a(b-c)\ge d(b+c).$ However, this implies $b>c$ and $a>d,$ which fails.

If $a,b$ or $c,d$ share any prime factors with each other then the result is obvious. Assume everything is relatively prime and take some $p \mid e $ so that $a^4 \equiv -b^4 \pmod{ p^5 }, $ or $(ab^{-1})^8 \equiv 1 \pmod{ p^5 }.$ Then $8 \mid p^4 (p-1) $, so $p=2$ or $p \equiv 1 \pmod {8} $. When $2\mid e $ we have an easy time, and considering only $e $s with odd prime factors leads to $e^5 \equiv 1\pmod {8}$, so one of $a,b $ and $c,d $ is even. The only case I couldn't resolve was $a,c$ both even. I also noted that 2 is a quadratic residue for the $p \equiv 1 \pmod {8}, $ but I don't expect that to be useful. Maybe 1 or 2 points? Looks like I missed the main factorization though.

Iggy Iguana wrote: One approach is to use the fact that $ac+bd$ divides $$a^4c^4-b^4d^4=(e^5-b^4)c^4-b^4(e^5-c^4)=e^5(c^4-b^4).$$ Now we get that if $ac+bd$ is prime, then $ac+bd$ must divide at least one of $e, b+c, b-c, b^2+c^2.$ The first four can be proven not true by not hard methods... Yes, that works... to clear up, the "not hard" methods are just the observation that $p = ac+bd$ is larger than each of $e$, $b+c$, $b-c$. So the fifth power was just there for size reasons so much trying to do number theory in $\mathbb Z[\zeta_8]$.

robinpark wrote: AkshajK says he solved it using the observation $(a^2-d^2)c^2 - (b^2 - c^2)d^2 = (ac+bd)(ac-bd)$ or something idk apparently it factors nicely Wait no it was the same factorization iggy found

Suppose $ac+bd$ is not composite. Clearly $ac+bd>2$ and $ad+bc>2$, so $ac+bd$ is an odd prime. Then since $a,b$ and $c,d$ are symmetric pairs of variables, WLOG $ad+bc$ is also an odd prime. Thus $(ac+bd)+(ad+bc)=(a+b)(c+d)$ is even, so either $2\mid(a+b)$ or $2\mid(c+d)$. WLOG let $2\mid(a+b)$. So either $a,b$ are both even or both odd. If $a,b$ are both even, then $2\mid ac$ and $2\mid bd$, so $2|(ac+bd)$, and since $ac+bd>2$, $ac+bd$ is composite, contradiction. If $a,b$ are both odd, then $a^4\equiv b^4\equiv1\pmod{16}$, so $e^5\equiv a^4+b^4\equiv2\pmod{16}$. However, this is impossible, since $e^5\equiv0\pmod{16}$ for even $e$ and $e^5\equiv e\pmod{16}$ for odd $e$. Thus we have a contradiction. Thus $ac+bd$ is composite. Is there a hole in this proof?

stephcurry wrote: Oh wow if you didn't solve it then maybe nobody did (or very few ) Ah, I'm not invincible or anything Sounds like plenty of my friends got this one today. program4 wrote: WLOG $ad+bc$ is also an odd prime. You can't do the WLOG step here. You're told that $ac+bd$ is a prime, but nothing about $ad+bc$.

couldn't find the desired factorization

v_Enhance wrote: program4 wrote: Suppose $ac+bd$ is not composite. Clearly $ac+bd>2$ and $ad+bc>2$, so $ac+bd$ is an odd prime. Then since $a,b$ and $c,d$ are symmetric pairs of variables, WLOG $ad+bc$ is also an odd prime. You can't do the WLOG step here. You're told that $ac+bd$ is a prime, but nothing about $ad+bc$. Thanks, but I'm still confused about why this is not allowed. So suppose that $ac+bd$ is an odd prime. Why can't we redefine the problem using the transformation $(a,b,c,d)\rightarrow(a,b,d,c)$? The corresponding equation is still $a^4+b^4=d^4+c^4=e^5$, so why can't we say $ad+bc$ is also an odd prime in the transformed problem (which is equivalent to the original)? Another way to put it, if $ac+bd$ is composite, then shouldn't $ad+bc$ be necessarily composite, and vice versa - aren't these two statements equivalent since $c$ and $d$ play exactly symmetric roles in the equation? Let's say we actually found a solution $(a,b,c,d)$ such that $ac+bd$ is composite. This solution satisfies both $a^4+b^4=c^4+d^4=e^5$ and $a^4+b^4=d^4+c^4=e^5$, so shouldn't $(a,b,d,c)$ also be a solution, in which case $ad+bc$ is also composite?

program4 wrote: v_Enhance wrote: program4 wrote: Suppose $ac+bd$ is not composite. Clearly $ac+bd>2$ and $ad+bc>2$, so $ac+bd$ is an odd prime. Then since $a,b$ and $c,d$ are symmetric pairs of variables, WLOG $ad+bc$ is also an odd prime. You can't do the WLOG step here. You're told that $ac+bd$ is a prime, but nothing about $ad+bc$. Thanks, but I'm still confused about why this is not allowed. So suppose that $ac+bd$ is an odd prime. Why can't we redefine the problem using the transformation $(a,b,c,d)\rightarrow(a,b,d,c)$? The corresponding equation is still $a^4+b^4=d^4+c^4=e^5$, so why can't we say $ad+bc$ is also an odd prime in the transformed problem (which is equivalent to the original)? Another way to put it, if $ac+bd$ is composite, then shouldn't $ad+bc$ be necessarily composite, and vice versa - aren't these two statements equivalent since $c$ and $d$ play exactly symmetric roles in the equation? Let's say we actually found a solution $(a,b,c,d)$ such that $ac+bd$ is composite. This solution satisfies both $a^4+b^4=c^4+d^4=e^5$ and $a^4+b^4=d^4+c^4=e^5$, so shouldn't $(a,b,d,c)$ also be a solution, in which case $ad+bc$ is also composite? I believe it's because $ad+bc$ has to be composite for all possibilities.

Basically what you proved is that at least one of $ac+bd$ and $ad+bc$ is composite. You need to show that both must be composite to solve the problem.

program4 wrote: So suppose that $ac+bd$ is an odd prime. Why can't we redefine the problem using the transformation $(a,b,c,d)\rightarrow(a,b,d,c)$? The corresponding equation is still $a^4+b^4=d^4+c^4=e^5$, so why can't we say $ad+bc$ is also an odd prime in the transformed problem (which is equivalent to the original)? Another way to put it, if $ac+bd$ is composite, then shouldn't $ad+bc$ be necessarily composite, and vice versa - aren't these two statements equivalent since $c$ and $d$ play exactly symmetric roles in the equation? This is circular reasoning. To give an example of the wrong logic, considering: Bogus problem: Prove that if $x+y = 1$ then $3x+y \ge 2$. Bogus proof: Assume for contradiction not, so $3x+y < 2$. "Without loss of generality", $x+3y < 2$ as well. Adding gives $4(x+y) < 4$, which is a contradiction. You're making the same error as here.

Ah okay, thanks everyone for the clarification.

Hi this was the hardest problem I've ever seen also e^5 is pretty bad and silly anyway can someone/people please check over this to see if it's right (I literally just typed exactly what I had on my paper) I'm the least sure about case 3 thanks

Rewritten solution. The key factorization is \[(a^4+b^4)c^2d^2-(c^4+d^4)a^2b^2=(ac+bd)(ac-bd)(ad+bc)(ad-bc)=e^5(cd-ab)(cd+ab)\]Let $ac+bd=p$, where $p$ is a prime. Case 1: $(ac+bd)(ac-bd)(ad+bc)(ad-bc)=e^5(cd-ab)(cd+ab)=0$ Then either $ac=bd$ or $ad=bc$, and $ab=cd$. If $ac=bd$, then since $a,b,c,d$ are all distinct, $ac+bd$ is not a prime. If $ad=bc$ and $ab=cd$, then $a(d-b)=c(d-b)\implies a=c$, a contradiction. Case 2: $p\mid e$. Let $n=\nu_p(e)$. We have $p^{5n-1}\mid (ad+bc)(ad-bc)$. Now we use the fact that \[p^2=(ac+bd)^2=a^2c^2+2abcd+b^2d^2>ad+bc.\] This implies $\nu_p((ad+bc)(ad-bc))<4$, contradiction. Case 3: $p\nmid e$ and $e^5(cd-ab)(cd+ab)\ne 0$. In this case $p\mid cd+ab$ or $p\mid cd-ab$. Both imply $ac+bd\le cd+ab$. So $c(a-d)\le b(a-d)$. Thus, either $b> c$ and $a>d$ or $c> b$ and $d> a$. Both of these imply $a^4+b^4\ne c^4+d^4$, a contradiction.

Suppose that $p=ac+bd$. Then, we have $p\mid a^4c^4-b^4d^4$, so since $a^4=e^5-b^4$ and $c^4=e^5-d^4$, we have $a^4c^4=e^5(e^5-b^4-c^4)+b^4d^4$, which means that $p\mid e^5(e^5-b^4-c^4)$. Therefore, either $p\mid e$ or $p\mid e^5-b^4-c^4$. Suppose that $p\nmid e$. Since $a^4b^4+b^8=b^4c^4+b^4d^4\equiv b^4c^4+a^4c^4\pmod p$, we have $p\mid(b^4+a^4)(b^4-c^4)=e^5(b^4-c^4)$. This means that $b^4\equiv c^4\pmod p$ and $b^4+c^4\equiv e^5\pmod p$, which means that $a^4\equiv b^4\equiv c^4\equiv d^4\pmod p$. However, since $a$, $b$, $c$, and $d$ are distinct and less than $p$, we must have that $a+b+c+d=2p$, which is impossible since $2ac>a+c$ and $2bd>b+d$. Therefore, we must have $p\mid e$, so $$a^4\equiv-b^4\pmod{p^5}$$and $$c^4\equiv-d^4\pmod{p^5}.$$However, this means that $p^5\mid(ac-bd)(ac+bd)(a^2c^2+b^2d^2)$, which is impossible since $ac-bd<p$, $ac+bd=p$, and $a^2c^2+b^2d^2<p^2$. Therefore, $ac+bd$ must be composite.

Let $p=ac+bd$, $p$ is a prime, then $d=\frac{p-ac}{b}$. $\therefore a^4+b^4=c^4+\left(\frac{p-ac}{b}\right)^4\Rightarrow b^4(a^4+b^4)=b^4c^4+(p-ac)^4=b^4c^4+a^4c^4-4p^3ac+6p^2a^2c^2-4pa^3c^3+p^4$. $\therefore (b^4-c^4)(a^4+b^4)=p(p^3-4acp^2+6a^2c^2p-4a^3c^3)\text{ }\cdots (*)\Rightarrow p\mid (b^4-c^4)(a^4+b^4)=(a^4+b^4)(b-c)(b+c)(b^2+c^2)$. $p=ac+bd>b+c,|b-c|\Rightarrow p\mid (a^4+b^4)(b^2+c^2)\Rightarrow p\mid (a^4+b^4)$ or $p\mid (b^2+c^2)$ If $p\mid (a^4+b^4)=e^5\Rightarrow p|e\Rightarrow p^5\mid e^5=a^4+b^4$. From $(*)$, $p^4\mid (p^3-4acp^2+6a^2c^2p-4a^3c^3)\Rightarrow p\mid 4a^3c^3$ which is not true. If $p\mid (b^2+c^2)$, $(a^2+d^2)(b^2+c^2)=(ac+bd)^2+(ab-cd)^2=p^2+(ab=cd)^2\Rightarrow p\mid (ab-cd)$. If $ab-cd=0\Rightarrow a^4b^4=c^4d^4$, $a^4+b^4=c^4+d^4\Rightarrow \{a,b\}=\{c,d\}$ which is not true. If $|ab-cd|\leqslant p$, let $ab-cd\leqslant p=ac+bd\Rightarrow ab>ac,bd\Rightarrow b>c,a>d\Rightarrow a^4+b^4>c^4+d^4$ which is not true.

Quite nice, actually; half of the problem's difficulty is how scary it looks. Set $p = ac+bd$, and assume first that $p \nmid e$. Observe that for size reasons, we cannot have $p \mid a$ or any of the other three. Using this, note that \begin{align*} a^4+b^4 &\equiv \left(\frac{bd}a\right)^4 + \left(\frac{bd}b\right)^4 \\ &\equiv (bd)^4 \cdot \frac{a^4+b^4}{a^4b^4} \\ 1 &\equiv \frac{b^4}{a^4} \pmod p. \end{align*}Then $$p \mid a^4-b^4=(a-b)(a+b)(a^2+b^2).$$Similarly, $p \mid c^4 - b^4$. However, clearly we cannot have $p \mid a-b$ or $p \mid a+b$, and furthermore $$(ac+bd)^2 \leq (a^2+b^2)(c^2+d^2),$$so the only possibility is for $a^2+b^2 = c^2+d^2 = p$. This implies $\frac ac = k = \frac bd$. Yet by the original equality, this implies $k=1$, which is a distinctness contradiction. Thus, we have a contradiction in this case. Finally, if $p \mid e$, then $(ac+bd)^5 \mid a^4+b^4$ which is clearly impossible.

Assume the contrary, Let $p = ac + bd$ be a prime number. Then we have $d \equiv -\frac{ac}{b} (p)$, so $a^4 + b^4 \equiv c^4 + \frac{a^4c^4}{b^4} (p)$, which is equivalent to $(a^4 + b^4)(b^4 - c^4) \equiv 0 (p)$. Assume $p \mid b^4 - c^4$. Then $p \mid (b+c)(b-c)(b^2 + c^2)$, so $p \mid b^2 + c^2$, otherwise $ac + bd = p \le b+c$, a contradiction. Note that $\frac{b}{c} \equiv -\frac{a}{d} (p)$, so we get $-1 \equiv \frac{b^2}{c^2} \equiv \frac{a^2}{d^2} (p)$. Hence $p \mid a^2 + d^2$. If $c < a$, then we have $c^2 + bd < ac + bd = p \le b^2 + c^2$, thus $d < b$. But this contradicts $a^4 + b^4 = c^4 + d^4$. Otherwise $a^2 + bd < ac + bd = p \le a^2 + d^2$, so $b < d$, contradicting the fact that $a^4 + b^4 = c^4 + d^4$. Hence $p \nmid b^4 - c^4$. So $p \mid a^4 + b^4$, which means $p \mid e$. Hence $p^5 \mid e^5 = a^4 + b^4$, so $a^5 + b^5 < (ac + bd)^5 \le e^5 = a^4 + b^4$, a contradiction. Thus we're done. $\blacksquare$

buh had to use a bunch of hints . nt bad Let $p=ac+bd$ be prime, clearly $p\ne 2.$ Notice $(a^4+b^4+c^4+d^4)(a^4-b^4+c^4-d^4)=(a^4+b^4)(a^4-b^4)+(c^4+d^4)(c^4-d^2)+2p(ac-bd)(a^2c^2+b^2d^2)=\frac12(a^4+b^4+c^4+d^2)(a^4-b^4+c^4-d^4).$ Thus $p\mid (a^4+b^4+c^4+d^4)(a^4-b^4+c^4-d^4)=4(a^4+b^4)(a^4-d^4)=4e^5(a-d)(a+d)(a^2+d^2).$ Now $(ac+bd)^5>(a+d)^5>a^4+d^4$ means $p>e$ and clearly $p>a-d,a+d$ so $p\mid a^2+d^2$ and similarly $p\mid b^2+c^2.$ WLOG assume $a^2+d^2\ge c^2+b^2,$ then from $a^4-d^4=c^4-b^4$ we have $a^2-d^2\le c^2-b^2$ so $d>b,a>c$ and $p=ac+bd>b^2+c^2,$ contradiction.

Let $p=ac+bd$ be a prime. For size reasons, $p$ cannot divide any of $a,b,c,d$. Furthermore, we show that $p>e$, so $p \nmid e$. Note that \[e^5 = a^4+b^4 \le a^5+b^5 < (ac+bd)^5 = p^5 \implies p>e.\] Now, take everything modulo $p$. We have \begin{align*} &\phantom{\implies} ac \equiv -bd \\ &\implies a^4c^4 \equiv b^4d^4 \\ &\implies a^4(e^5-d^4) \equiv (e^5-a^4)d^4 \\ &\implies a^4e^5-a^4d^4 \equiv d^4e^5-a^4d^4 \\ &\implies e^5(a^4-d^4) \equiv e^5(a-d)(a+d)(a^2+d^2) \equiv 0 \pmod{p}. \end{align*} This implies $p \mid a^2+d^2$ and we similarly have $p \mid b^2+c^2$. Hence, \[a^2+d^2 \ge p\] and \[b^2+c^2 \ge p.\] These give us the following system of equations: $$ \begin{cases} a(a-c) &\ge d(b-d) \\ c(c-a) &\ge b(d-b) \end{cases} $$ If $a>c$, we have $d>b$, and if $c>a$, we have $b>d$, upon which both cases violate one of the inequalities, a contradiction. $\square$

very nice and easy nt!! somewhat similar to inmo 2023 p3? WLOG, say $a < c$. SFTSOC, $ac + bd = p$, where $p$ is a prime. This means $ac \equiv -bd\mod{p} \implies a^4c^4 \equiv b^4d^4\mod{p}$. Note that $c^4 = e^5 - d^4$, $b^4 = e^5 - a^4 \implies a^4(e^5 - d^4) \equiv d^4(e^5 - a^4) \mod{p}$. This means that: $$p \mid a^4(e^5 - d^4) - d^4(e^5 - a^4) = e^5(a^4 - d^4) = e^5(a - d)(a + d)(a^2 + d^2)$$ If $p \mid e^5$, we have $p \mid e \implies p^5 \mid e^5 \implies p^5 \mid a^4 + b^4$. Notice that $p^5 = (ac + bd)^5 \implies a^4 + b^4 \geq (ac + bd)^5$, contradiction as both $c$ and $d$ can't equal $1$. If $p \mid a - d$, we have one of $a, d > p \implies ac + bd > p$, contradiction. If $p \mid a + d$, we have $a + d \geq p \implies ac + bd > p$, contradiction. If $p \mid a^2 + d^2$, we have $a^2 + d^2 \geq ac + bd \implies d(d - b) \geq a(c - a)$, and since $c > a$, we have $d > b$ as well, meaning $c^4 + d^4 > a^4 + b^4$, contradiction. Hence we are done.

FTSOC $p = ac + bd$ for some prime $p$. Then clearly by size $ac + bd$ doesn't divide $ a, b, c, d, e$. So then take expressions modulo $p$. We have $(ac)^4 \equiv (bd)^4 \pmod p \implies a^4(e^5 - d^4) \equiv d^4(e^5 - a^4) \pmod p \implies e^5(a^4 - d^4) \equiv e^5(a - d)(a + d)(a^2 + d^2) \equiv 0 \pmod p$. Then by size we have $ac + bd \mid a^2 + d^2$ and analogously $ac + bd \mid b^2 + c^2$. However this implies that $a^2 + d^2 \geq ac + bd \implies a(a - c) + d(b - d) \geq 0$ and $c(c - a) + b(d - b) \geq 0$. Since $a, b, c, d$ are distinct we must have either $a > c, b < d$ or $a < c, b > d$. Considering either of them in one equation leads to a contradiction in the other equation.

Let $a$, $b$, $c$, $d$ and $e$ be distinct positive integers such that, $$a^4 + b^4 = c^4 + d^4 = e^5$$Prove that $ac + bd $ is composite. Proof. Assume otherwise, and let $p = ac + bd$ be a prime. Then standard manipulations give, \begin{align*} a^4 + b^4 &\equiv c^4 + d^4 \pmod{p}\\ \iff \left(-\frac{bd}{c}\right)^4 + b^4 &\equiv c^4 + d^4 \pmod{p}\\ \iff (b - c)(b + c)(b^2 + c^2)(c^4 + d^4) &\equiv 0 \pmod{p} \end{align*}Clearly $p \nmid b - c$, $p \nmid b + c$ due to size reasons. Thus we have reduced to, \begin{align*} e^5(a^2 + d^2) \equiv 0 \pmod{p} \end{align*}If $p \mid a^2 + d^2$ then clearly $p \mid b^2 + c^2$. Then due to size reasons we require $a > c$, $b > d$ or $a < c$, $b < d$, as if $a > c$ and $b < d$ then $ac + bd \nmid b^2 + c^2$ as $a c > c^2$ and $bd > c^2$. However in either of these cases, without loss of generality say $a > c$ and $b > d$, we find $a^4 + b^4 > c^4 + d^4$ a contradiction. Then the final case is $p \mid e^5$. Clearly this would force $p \mid e$ as by assumption $p$ is prime, and hence we would find $p^5 \mid a^4 + b^4, c^4 + d^4$ which fails due to size reasons once more.

Usamo 2015 Here is a sketch Let $ac+bd =p $ be a prime. Firstly assume WLOG $a>c$ then we have that \[p | e^5(a-d)(a+d)\left(a^2+d^2\right).\] Now we can see that $p>e$ and $p>a+d$. Hence $p| a^2+d^2$, so $ac+bd\le a^2+d^2$ hence we have $a^4 + b^4 > c^4 + d^4$, which is a contradiction.

Suppose $ac+bd$ is prime, we have that $p\mid a^4c^4-b^4d^4=e^5(c-b)(c+b)(c^2+b^2)$, thus we get $p\mid c^2+b^2>ac+bd$ implying $a>c$ and $b>d$ or opposite. Thus $a^4+b^4>c^4+d^4$ which is a contradiction. This is a terrible ugly problem.

Haven't seen this solution in forum yet so might be wrong: Took like 3 hours of staring at the problem not knowing what to do, but after trying to use the $a^4+b^4=c^4+d^4$ condition artificially stuff worked out. Note: now looking at it, assuming cd-ab isn't zero means this is slight fakesolve, but that is easily resolved and it seems otherwise this solution is probably equivalent to others . Assume FSOC that $ac + bd$ is prime. Then, $ac+bd \mid (ac+bd)(a^3d + b^3c) $ $= cd(a^4 + b^4) + ab(b^2c^2+a^2d^2)$ (1) and $ac+bd \mid (ac+bd)(a^3d + b^3c + c^3b + d^3a) $ $= cd(a^4 + b^4) + ab(b^2c^2+a^2d^2) + ab(c^4 + d^4) + cd(b^2c^2 + a^2d^2)$ $ = (ab+cd)(a^4+b^4)+(ab+cd)(b^2c^2+a^2d^2) $ $= (ab+cd)(a^4+b^4+b^2c^2+a^2d^2)$. As $ac+bd > ab + cd$ by rearrangement, thus $ac + bd \mid (a^4+b^4+b^2c^2+a^2d^2) \mid cd(a^4+b^4) + cd(b^2c^2+a^2d^2)$ (2) Subtracting (1) and (2), we have that: $ac + bd \mid (cd-ab)(b^2c^2+a^2d^2)$, and again by size, $ac + bd \mid b^2c^2+a^2d^2 \implies ac+bd \mid a^4+b^4 = e^5$ But $ac+bd < e$, contradiction.

Please contact westskigamer@gmail.com if there is an error with my solution for cash bounties by 3/18/2025.

Q. a^4 + b^4 = c^4 + d^4 = e^5. Show that ac + bd is composite. Suppose ac + bd = p is a prime. Then, (a-d)(b-c) = ab + cd - ac - bd. (a+c)^2 + (b+d)^2 - (a-c)^2 - (b-d)^2 = 4p. Suppose q\mid e. [q=2 is clearly impossible] Then, none of a, b, c, d are divisible. So, q \mid (a/b)^4 + 1. So, q = 1 (mod 8). Suppose e is even. Then, a, b, c, d, all odd == not possible mod 4. Hence, e is odd. So, say a is even. Then, c must be even. Note, x^4 = 0, 1 (mod 16). Hence, e^5 = 1 (mod 16). Then, e = 1 (mod 16). (a^2-c^2)(a^2+c^2)=(d^2-b^2)(d^2+b^2) ac + bd | (a^4c^4 - b^4d^4) = a^4(e^5-d^4) - (e^5-a^4)d^4 = e^5 (a^4 - d^4) = e^5 (c^4 - b^4). p | e^5 then p^5 | e^5 so (ac + bd)^5 <= |a^4c^4 - b^4d^4| which is obviously false. Hence, p | a^4 - d^4. So, either p | a-d [false], or p | a+d [also false] or p | a^2 + d^2. Similarly, p | b^2 + c^2. Note, if a>c then b<d. So, p = ac + bd > c^2 + b^2, the required contradiction. $\blacksquare$