Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.)

share|improve this question
    
see en.wikipedia.org/wiki/Beal%27s_conjecture I see this, you get $m | (m+r)$ and $n | (1+r)$ –  Will Jagy Jan 24 '13 at 20:15
    
There is a slight variant possible if your $f$ is already divisible by an $n$-th power –  Will Jagy Jan 24 '13 at 20:32
    
Yes, Adam, the method works in general as you have stated it. –  Gerry Myerson Jan 25 '13 at 11:59
    
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 –  Gerry Myerson Jan 25 '13 at 12:06
    
@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 $g^r$. –  Adam Bailey Jan 25 '13 at 21:20

2 Answers 2

up vote 1 down vote accepted

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.)

share|improve this answer

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.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.