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Does the equation $a^{2} + b^{7} + c^{13} + d^{14} = e^{15}$ have a solution in positive integers

Like FLT -- cannot see how to attack this one :(

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    do the integers need to be distinct? – 2017-02-26
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    I'm assuming they don't have to be as otherwise it would have been stated .This is the exact wording of the problem when I found it ..Just can't solve it ! – 2017-02-26
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    Well, fortunately it is not as difficult "Like FLT". – 2017-02-26
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    This question has been answered, but I think the following question needs to be asked, “does this equation have any smaller solutions?” Although I’m perfectly happy to post it, I feel @Randin Michael Divelbiss should have the opportunity first. – 2017-02-28
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    @OldPeter: There is a smaller solution since $x=u/2$ works as well and the conditional equation is now $182u+1=15y$. However, I'm wondering if there is a solution where at least one of $a,b,c,d$ is not evenly divisible by $4$. – 2017-03-01
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    @Tito Piezas III: I think I've found a way, but it's too big for a comment. – 2017-03-01

3 Answers 3

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Given, $$a^{2} + b^{7} + c^{13} + d^{14} = e^{15}\tag1$$

Assume $a=4^{182x},\;b=4^{52x},\;c=4^{28x},\;d=4^{26x},\;e=4^{y}$. Substituting into $(1)$ we get, $$4^{364x+1}=4^{15y}\tag2$$

which is true if, $$364x+1=15y\tag3$$

The last equation has solutions of the form

$$x=11+15n\\y=267+364n$$

Thus the equation $(1)$ has infinitely many solutions, the simplest one using $n=0$,

$$(a,b,c,d,e)=(4^{2002},4^{572},4^{308},4^{286},4^{267})$$

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    I missed a minus sign, now you can try and check. Its absolutely fine – 2017-02-26
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    +1 Very clever. (I made some formatting changes. I hope it is ok.) – 2017-02-26
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    Thanks and yes the formatting makes it look better :-) – 2017-02-26
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    Nice answer! I wonder if there are other solutions... – 2017-02-27
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    I think there are , but I haven't thought of such a substitution yet. – 2017-02-27
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Please be kind with my attempt at a solution, and let me know where I’ve gone wrong.

Using the method shown by @Aditya Narayan Sharma, $$a^{2} + b^{7} + c^{13} + d^{14} = e^{15}\tag1$$

Put $a=2^{91x},\;b=2^{26x},\;c=2^{14x},\;d=2^{13x},\;e=2^{y}$. Then (1) becomes, $$2^{182x}+2^{182x}+2^{182x}+2^{182x}=2^{15y}$$ $$4*2^{182x}=2^{15y}$$ $$2^{182x+2}=2^{15y}\tag2$$ So, $$182x+2=15y$$ $$182x-15y=-2\tag3$$ Giving, $$x=14+15n$$ $$y=170+182n$$

Using $n=0$,

$$(a,b,c,d,e)=(2^{1274},2^{364},2^{196},2^{182},2^{170})$$

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Just in case anybody’s considering a brute force approach, if we allow zero for $(a,b,c,d)$ there are $36$ solutions within $7.9E+28$.

$$(a,b,c,d,e)=(p^{15},0,0,0,p^2)\tag1$$

$$(a,b,c,d,e)=(pq^7,q^2,0,0,q):q=p^2+1\tag2$$

$$(a,b,c,d,e)=(pq^7,0,0,q,q):q=p^2+1\tag3$$

$$(a,b,c,d,e)=(pr^7,r^2,0,r,r):r=p^2+2\tag4$$

Updated 3 March 2017 to correct typo in $a$ values.

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    "Positive integers " only . 0 not allowed – 2017-03-02
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    @Aditya Narayan Sharma. Yes, indeed, these results are outside of the specification of the question. However, running a program for an hour without producing any results always leaves questions about the quality of both the coding and testing, so I designed to produce some results with minimal extra run time. I’ve posted my data only to be helpful to others taking this approach. – 2017-03-02
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    +1 Yes of course,I myself love coding in python and running scripts to check solutions. You may also try to exclude zero and get some other bunch of identities. – 2017-03-02
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    @Aditya Narayan Sharma. I use VB6, which limits me to about $7.92281625142643E+28$ for exact arithmetic, unless I take masses of time producing non-standard, convoluted code. I’ve already covered all the possibilities within this range, but that only covers up to $e=84$. The program produced $36$ numerical solutions, from which I constructed the parametric “almost solutions”. – 2017-03-02
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    You can also search for polynomial that satisfy the equation. – 2017-03-03
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    Consider the Lowest Common Multiple of 2, 7, 13 and 14:     LCM(2, 7, 13, 14) = 13×14 = 182  And     182 = 15×12 + 2     182 ≡ 2 [15]  We want a multiple of 182 that is equal to 14 modulo 15 so:     7×182 = 1274 = 15×84 + 14 ≡ 14 [15]  Thus setting     ² = 4¹²⁷⁴ → = 4⁶³⁷     ⁷ = 4¹²⁷⁴ → = 4¹⁸²     ¹³ = 4¹²⁷⁴ → = 4⁹⁸     ¹⁴ = 4¹²⁷⁴ → = 4⁹¹  would yield:     ² + ⁷ + ¹³ + ¹⁴ = 4¹²⁷⁴ + 4¹²⁷⁴ + 4¹²⁷⁴ + 4¹²⁷⁴ = 4×4¹²⁷⁴ = 4¹²⁷⁵ = ¹⁵  with     ¹⁵ = 4¹²⁷⁵ → = 4⁸⁵  So ,       YES there is indeed positive integers , , , and such as:     ² + ⁷ + ¹³ + – 2017-03-05
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    +d^14 = ¹⁵  with     (, , , , ) = (4⁶³⁷, 4¹⁸², 4⁹⁸, 4⁹¹, 4⁸⁵)  being an example of such integers. – 2017-03-05
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    Indeed a fascinating problem. This solution is identical to my solution within my answer (the one above this answer), except I used powers of $2$ rather than $4$. In my comment under the original question I suggested “This question has been answered, but I think the following question needs to be asked, “does this equation have any smaller solutions?” Although I’m perfectly happy to post it, I feel @Randin Michael Divelbiss should have the opportunity first.” I just feel there should be a much smaller solution. – 2017-03-05
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    I would if I could Peter I like your solution that is more general . – 2017-03-06