# $x^3+y^4=7$ has no integer solutions

I am trying to prove that $x^3+y^4=7$ has no integer solutions, but i have no idea how to start, please helps. I have tried to consider mod 7 to restrict the number of possible $x^3$ because $x^3 \equiv -1,0,1 \pmod{7}$, but it is not working.

• I would try reducing mod some other numbers (not necessarily primes). 7 is not going to work. Commented Nov 27, 2013 at 12:52
• if there's a solution then x must be negative. Commented Nov 27, 2013 at 12:59
• Commented Nov 27, 2013 at 15:38

Consider the equation modulo $13$. Then $x^3$ can be $0,1,5,8,12$ and $y^4$ can be $0,1,3,9$. None of these add to $7$ modulo $13$.

I chose $13$ because $3|\phi(13)$ and $4|\phi(13)$, so I could get restrictions on both $x^3$ and $y^4$.

• Why should we in general consider multiples of $\phi(n)$? Is there some kind of theory behind this? What does guarantee me that if I consider all remainders of $x^k$ modulo $n$ where $k \mid \phi(n)$ the variety won't be that big?
– sve
Commented Nov 27, 2013 at 13:59
• Well, the principle is that you want $\mathrm{gcd}(k,\phi(n))$ to be large relative to $\phi(n)$. To avoid complicating things, let's take the case when $n$ is prime. Then we know that there are $\mathrm{gcd}(n-1,k)$ solutions to $x^k \equiv 1 \pmod{n}$ and hence only $1 + \frac{n-1}{\mathrm{gcd}(n-1,k)}$ different values that $x^k$ can take modulo $n$. Commented Nov 27, 2013 at 14:26
• Why there are exactly $gcd(n-1,k)$ solutions to the modulo equation? I fail to see this.
– sve
Commented Nov 27, 2013 at 14:35
• For prime $n$, take a primitive root $q$ modulo $n$. Then $q^{s\frac{n-1}{\mathrm{gcd}(n-1,k)}}$ is a solution by Fermat's little theorem, and these are different for $0\leq s < \mathrm{gcd}(n-1,k)$ since $q$ is a primitive root. Commented Nov 27, 2013 at 14:43
• And now that we know that $x^k \equiv 1$ (mod $n$) has $gcd(n-1, k)$ different solutions how can we tell that $x^k$ would take $1 + \frac{n-1}{gcd(n-1, k)}$ different values? Can't make the connection.
– sve
Commented Nov 27, 2013 at 14:59

universalset's answer is right, in these problems a useful heuristic is to find a number that produces remainders such that both sides cannot become equal. Here is a short Python program to find such numbers for this problem.

for m in range(2,100): #We hope to find such number less than 100
a= set([x**3%m for x in range(0,1000)]) #Hoping remainders will cycle somewhere less that 1000
b= set([y**4%m for y in range(0,1000)]) #Hoping remainders will cycle somewhere less that 1000

flag = 0;
for x in a:
if (7%m+x%m)%m in b:
flag=1;
break;

if flag==0:
print (m,a,b)