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Prove (elementary, meaning no high level theorems used) that there can not exist 4 prime numbers a,b,c,d $\geq$ 7 such that

\begin{equation}a^4+b^4+c^4+d^4=2^{2011}\end{equation}

I tried the following: The last digit l(d) of the fourth power of a prime number is 1 or 5. The last digit of $2^{2011}$ is 8. But there exists a combination that holds true:

\begin{equation}l(a)=1,l(b)=1,l(c)=1,l(d)=5\end{equation}

\begin{equation}1+1+1+5=8\end{equation}

Alexandru
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    Remember the primes are all greater than $5$ so none of them end in $5$. – Empy2 Jul 15 '15 at 13:09
  • For odd $a, a^4\equiv1\pmod{16}\implies \sum a^4\equiv4\pmod{16}$ But $2^{2011}\equiv0$ – lab bhattacharjee Jul 15 '15 at 13:10
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    We do not find any primes greater than 5 that have last digit 5 when taken to the fourth power, thus the only last digit possible is 1 and thus it is impossible to find such $a,b,c,d$. – Jan Jul 15 '15 at 13:13
  • Yes Michael and Jan I didn't see that you are right , the answer I was looking for – Alexandru Jul 15 '15 at 13:15
  • with a comma between that and you , above – Alexandru Jul 27 '15 at 17:58
  • For all prime numbers $p$ such that $p\neq 3$, we have $p^4\equiv 1 \pmod 3$ Therefore, for all prime numbers $a, b, c, d\geqslant 7$, one has $a^4 + b^4 + c^4 + d^4 \equiv 4 \pmod 3 \equiv 1 \pmod 3\Rightarrow 2^{2011}\equiv 1\pmod 3$. So now, subtract $1$ from $2^{2011}$ and then divide that result by $3$. If it is not an integer, then $2^{2011}\not\equiv 1 \pmod 3$ which is contrary to the foregoing. $$\therefore a^4 + b^4 + c^4 + d^4\neq 2^{2011}$$ This would also imply that $2^{2011}\equiv 2 \pmod 3$, if this is the case, since $\gcd(2, 3) = 1$. – Mr Pie Jan 09 '18 at 09:49

2 Answers2

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If $n$ is odd, $n^4\equiv 1\pmod{16}$, hence if $a,b,c,d$ are prime numbers and at least one of them is odd, $$ a^4+b^4+c^4+d^4 \in \{1,2,3,4\}\pmod{16}.$$ However, $2^{2011}\equiv 0\pmod{16}$ and $2^4+2^4+2^4+2^4$ is way smaller than $2^{2011}$.

Jack D'Aurizio
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Considering in mod $3$ helps. We have $(\pm 1)^4\equiv 1\pmod 3$, but $2^{2011}\equiv (-1)^{2011}\equiv -1\pmod 3$.

mathlove
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