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

Solution 1:

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 $n=0$,

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

Solution 2:

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 $n=0$,

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

Solution 3:

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.