I am considering the following Diophantine Equation - the approach I tried became the study of too many different cases - so many that I left it and tried to find an easier way. I wonder if anyone could shed any further light on the matter. The equation is interesting since it has 3 different exponents. $$ a^2 +b^3 = c^4 $$ $a$, $b$, $c$, are any integers. The approach I mentioned began by subtracting the $a^2$ on both sides and factoring the difference of two squares - but as I said, it quickly became very tedious.
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asked Jan 10, 2015 at 21:41
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$\begingroup$ What are we seeking? A solution where none of the variables is $0$? Infinitely many solutions of this type? All solutions? There are other possibilities. $\endgroup$– André NicolasJan 10, 2015 at 22:17
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$\begingroup$ All solutions. The case when one of the variables is 0 included, although these are trivial and easy to deduce. $\endgroup$– Asier CalbetJan 10, 2015 at 22:23
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$\begingroup$ see Lemma 3.2.7 p.22 here cecm.sfu.ca/~nbruin/thesis.pdf . Just set $(a,b,c)\rightarrow (a,-b,c)$ in the set of solutions. $\endgroup$– 111Jan 10, 2015 at 22:24
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$\begingroup$ Really, so one can obtain a non-trivial solution from those parametric solutions? $\endgroup$– Asier CalbetJan 10, 2015 at 22:26
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1$\begingroup$ I needed it,in my phd... $\endgroup$– 111Jan 10, 2015 at 22:48
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2 Answers
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Here's an infinite family of solutions,
$$\big(4q^2(p^2-2)\big)^2+(4dq^2)^3 = (2pq)^4$$
where $p,q$ solve the Pell equation $p^2-d^3q^2=1$. If you want the "reversed" form,
$$p^4 + (q^2-1)^3 = (q^3+q)^2$$
where $p^2-3q^2=1$. And an amazing one by Enrico Jabara,
$$(2^{88}7^{47}n^4)^3 + (2^{92}7^{49}n)^4 + (2^{32}7^{17}n^4)^5 + (2^8 7^4n^4)^7 + (n^4-2^{52}7^{28})^8 = (n^4+2^{52}7^{28})^8$$
which he probably found by expanding,
$$(an^4)^3 + (bn)^4 + (cn^4)^5 + (dn^4)^7 + (n^4-e)^8 = (n^4+e)^8$$
collecting powers of $n$, and solving for $a,b,c,d,e$.
answered Jan 11, 2015 at 0:03
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I thought it possible to write a solution without using equations Pell.
$$a^2+b^3=c^4$$
If you make this change.
$$p=tz(2zk^2+t)$$
$$s=tzk^2(2zk^2-t)$$
The result of such decision.
$$a=sp^3$$
$$b=2tzk^2p^2$$
$$c=kp^2$$
Where the number $t,z,k$ - integers and set us. You may need after you get the numbers, divided by the common divisor.
