Diophantine equation $a^m + b^m = c^n$ ($m, n$ coprime) 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.)
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 $2m^{-1} + n^{-1} &lt; 1$.

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

We can see there are primitive solutions if m &lt;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.