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Arising from this recent question, and in particular the answer by Gerry Myerson, it occurs to me that, if $m$ and $n$ are coprime integers, non-trivial solutions can be found to any Diophantine equation of this form (or with more powers of $m$ on the left hand side):

$$a^m + b^m = c^n$$

The method is to take any integers d and e, find $f = d^m + e^m$, then find an integer $r$ such that $r$ is congruent to $0 \mod m$ and to $-1 \mod n$. The Chinese Remainder Theorem tells us that such an integer can be found. Then:

$$(d^m)(f^r) + (e^m)(f^r) = f(f^r)$$

has powers of m on the left and a power of n on the right.

Is this correct, or am I missing something? (I realise that to find solutions with no common divisor is much harder or perhaps impossible.)

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  • $\begingroup$ see en.wikipedia.org/wiki/Beal%27s_conjecture I see this, you get $m | (m+r)$ and $n | (1+r)$ $\endgroup$
    – Will Jagy
    Jan 24, 2013 at 20:15
  • $\begingroup$ There is a slight variant possible if your $f$ is already divisible by an $n$-th power $\endgroup$
    – Will Jagy
    Jan 24, 2013 at 20:32
  • $\begingroup$ Yes, Adam, the method works in general as you have stated it. $\endgroup$
    – Gerry Myerson
    Jan 25, 2013 at 11:59
  • $\begingroup$ Only $10$ solutions are known of $x^p+y^q=z^r$ with $x$, $y$, $z$ coprime and $p^{-1}+q^{-1}+r^{-1}\lt1$. See towards the end of ams.org/notices/199711/beal.pdf $\endgroup$
    – Gerry Myerson
    Jan 25, 2013 at 12:06
  • $\begingroup$ @WillJagy,thank you, yes if $f$ happens to be divisible by an $n$-th power, say $f = gh^n$, then it suffices to multiply by $g^r$. $\endgroup$
    – Adam Bailey
    Jan 25, 2013 at 21:20

2 Answers 2

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This question has been answered in the above comments by Gerry Myerson and Will Jagy. The method in the question is correct. There is also a variant possible if $f$ is divisible by an $n$-th power. Solutions obtained by this method always have $a,b,c$ with a common divisor. It is much harder to find solutions with $a, b, c$ coprime, especially if it also required that $2m^{-1} + n^{-1} < 1$.

(This answer is posted according to guidance in this question on meta.)

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Have a look at this experimental result with Pari gp for $a^{m}+b^{m}+c^{m} = d^{m+1}$ and m =3. $? k=1000;for(a=1,k,for(b=a,k,for(c=b,k,if(ispower(a^3+b^3+c^3,4,&n)&gcd(a,b)==1& gcd(a,c)==1&gcd(b,c)==1,print([a,factor(a),b,factor(b),c,factor(c),n,f actor(n)]))))) [19, Mat([19, 1]), 89, Mat([89, 1]), 117, [3, 2; 13, 1], 39, [3, 1; 13, 1]] [75, [3, 1; 5, 2], 164, [2, 2; 41, 1], 293, Mat([293, 1]), 74, [2, 1; 37, 1]] [81, Mat([3, 4]), 167, Mat([167, 1]), 266, [2, 1; 7, 1; 19, 1], 70, [2, 1; 5, 1; 7, 1]] [107, Mat([107, 1]), 163, Mat([163, 1]), 171, [3, 2; 19, 1], 57, [3, 1; 19, 1]] [222, [2, 1; 3, 1; 37, 1], 263, Mat([263, 1]), 961, Mat([31, 2]), 174, [2, 1; 3, 1; 29, 1]] [225, [3, 2; 5, 2], 362, [2, 1; 181, 1], 407, [11, 1; 37, 1], 106, [2, 1; 53, 1]] [323, [17, 1; 19, 1], 333, [3, 2; 37, 1], 433, Mat([433, 1]), 111, [3, 1; 37, 1]] [397, Mat([397, 1]), 441, [3, 2; 7, 2], 683, Mat([683, 1]), 147, [3, 1; 7, 2]]$

We can see there are primitive solutions if m <4 because there are probably some hidden identities; i am myself skilless and have not enough mathematical knowledge to find them; but one seems to be $ e^{3}+ f^{3}+ 3^2g^{3} =3g^{4}$ The problem is to determine what e,f and g exactly are.

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