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find all primes p for which $x^4+4=py^4$ is solvable for integers.

My try: I started out with $p=2$ . LHS must be even, thus $(x^4+4)\mod(16)=4$ similarly

$(2y^4)\mod (16)=${$0,2$}. Thus for $p=2$ there are no solutions.

Now I tried factoring i.e $x^4+4=(x^2+2x+2)(x^2-2x+2)=py^4.$ I dont know what to do next

Any ideas??

Alessio K
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Albus Dumbledore
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  • Do you want to get one formula for all prime solutions? I mean if there are no restrictions for $x$ and $y$ other than being integers, then setting $y = 1$ should alone give infinitely many solutions for odd $x$, for which the LHS produces prime $p$. – Dávid Laczkó Sep 03 '20 at 17:21
  • @DávidLaczkó thre are no other restrictions Anser given is p=5 is the only possible prime – Albus Dumbledore Sep 03 '20 at 17:23
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    Fourth powers mod 5 are either 0 or 1. This gives either $py^4\equiv0\mod5,x^4\equiv1\mod5$ or $p\equiv 4\mod5,y^4\equiv1\mod5,x^4\equiv0\mod5$. – Moko19 Sep 03 '20 at 17:36
  • Trivially, p=5,x=1,y=1 is a solution – Moko19 Sep 03 '20 at 17:38

2 Answers2

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It is easy to see that $x$ is odd (otherwise $4$ divides $py^4$ and $x, y$ are both even, which gives contradiction mod $16$). Then the $\gcd$ of $x^2 + 2x + 2$ and $x^2 - 2x + 2$ is $1$. Therefore we must have \begin{eqnarray}x^2 + 2x + 2 &=& u^4\\x^2 - 2x + 2 &=& pv^4\end{eqnarray} or \begin{eqnarray}x^2 + 2x + 2 &=& pu^4\\x^2 - 2x + 2 &=& v^4.\end{eqnarray} In the first case, we have $(x^2 + 1)^2 + 1 = (u^2)^2$ which is impossible. In the second case, we have $(x^2 - 1)^2 + 1 = (v^2)^2$ which leads to $x = 1$, and hence $y = 1$ and $p = 5$.

WhatsUp
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@WhatsUp has the answer, but I wanted to show just for fun how to systematically derive the other case $(X,Y,P)=(0,\pm 1,4)$. It doesn't answer the question because $P=4$ is not prime, but to those who ask "How can we derive the other solution without some kind of trial and error?" well here it is.

Consider the equation $$(p^2-q^2)^2+(2pq)^2=(p^2+q^2)^2$$

Let $pq=4$ so that $$(p^2-q^2)^2+64=(p^2+q^2)^2$$

Let $p=x^2+y^2$ and $q=x^2-y^2$ so that $p^2-q^2=(2xy)^2$ and $p^2+q^2=2(x^2+y^2)^2$ then $$(2xy)^4+64=4(x^2+y^2)^4$$ or $$(xy)^4+4=\frac 14(x^2+y^2)^4$$

The coefficient is rational, but we want it to be an integer. The requirement is that $x^2+y^2$ is even. So let $x=2s$ and $y=2t$ so it follows $$(4st)^4+4=4(s^2+t^2)^4.$$

Remember the constraint $pq=4$. Since $p=x^2+y^2$ and $q=x^2-y^2$ then it follows $(s^2+t^2)(s^2-t^2)=1$. Since $x$ and $y$ are integers then $s$ and $t$ are integers and so $s^2\pm t^2$ are also integers. Therefore $s^2+t^2=s^2-t^2=1$ or $2s^2=2$ or $s=\pm 1$ and thus $t=0$.

Therefore we derive $(X,Y,P)=(0,\pm 1, 4)$.

Mr Pie
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  • interesting! solution, but why should we take x=2s y=2t why not x,y both be odd – Albus Dumbledore Sep 03 '20 at 18:14
  • @Quantum because that would mean $\frac 14(x^2+y^2)^4$ is even, but $(xy)^4+4$ odd, which is a contradiction. Thanks for the praise though :) – Mr Pie Sep 03 '20 at 18:28
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    yes thanks ,just curious how did you even think of starting with that initial form. It is ingenious! – Albus Dumbledore Sep 03 '20 at 18:30
  • @Quantum haha, thank you! I thought of it because I have been studying these types of equations myself in my own leisure, and generalising many similar identities. The fact that $4$ is a square number was more specifically the prompt :P – Mr Pie Sep 03 '20 at 18:34
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    one day i hope i can do something like this! – Albus Dumbledore Sep 03 '20 at 18:37
  • @Quantum if you're passionate enough, then it will happen. For me personally, I just fiddle around with numbers all the time for its own sake and eventually find substitutions that allow me to be a bit crafty. An example I like is simply in expanding $(a+b-c)^2$. If $c^2=a^2+b^2$ then $(a+b-c)^2=2(c-a)(c-b)$. Or finding other identities like $$(p+u)^3-(p+v)^3+(q+v)^3-(q+u)^3=3(p-q)(u-v)(p+q+u+v).$$ At first I got lucky, but then I started to notice a few symmetries and began building a good eye to spot them. Eventually you just acquire a knack for it, I suppose. – Mr Pie Sep 03 '20 at 18:59