Power sum of primes

In this discussion I will prove the following statement:

The equation p1n+p2n=p3np_1^n+p_2^n=p_3^n will have solutions where nZn \in Z and all of p1,p2,p3p_1,p_2,p_3 are primes if and only if n=1n=1.

Case 1 n>2n>2

No need to prove this case, because any equation of the form an+bn=cna^n+b^n=c^n where n>2;nZ;aZ;bZ;cZn>2;n\in Z;a\in Z; b\in Z;c\in Z don't have any solutions see Fermat's last theorem

Case 2 n=2n=2

Let us assume there exists primes p1,p2,p3p12+p22=p32p_1,p_2,p_3 | p_1^2+p_2^2=p_3^2 p12=p32p22=(p3p2)(p3+p2)p_1^2=p_3^2-p_2^2=(p_3-p_2)(p_3+p_2) Now pairwise factors of p12p_1^2 are (p12,1),(p1,p1)(p_1^2,1),(p_1,p_1) now if p3p2=p1p3p2=p1=p3+p2p2=0p_3-p_2=p_1\Rightarrow p_3-p_2=p_1=p_3+p_2\Rightarrow p_2=0 which is not possible p3p2=1;p3+p2=p12p3p2<p3+p2\therefore p_3-p_2=1;p_3+p_2=p_1^2 \because p_3-p_2<p_3+p_2 But this also leads to contradiction because according to it p3=p2+1p_3=p_2+1 which is not possible for p3>3p_3>3 (\because if p3p_3 is odd then p2p_2 will be divisible by 22 and vice versa)

Therefore no solution exists for p3>3p_3>3. If p3=3p2=2p_3=3\Rightarrow p_2=2 but p32p22=3222=5p12=5p1=5,5p_3^2-p_2^2=3^2-2^2=5 \Rightarrow p_1^2=5 \Rightarrow p_1=\sqrt{5},-\sqrt{5} which also leads to contradiction.

Therefore no solution exists for n=2n=2

Case 3 n=1n=1

Let us assume there exists primes p1,p2,p3p1+p2=p3p_1,p_2,p_3 | p_1+p_2=p_3. p1+p2=p3p_1+p_2=p_3 In the above equation, if all of p1,p2,p3p_1,p_2,p_3 are primes, then p3>p2;p3>p1p_3>p_2;p_3>p_1 therefore p3p_3 is greater than the 2nd2_{nd} prime p3>3\Rightarrow p_3>3.

p3\Rightarrow p_3 is odd \therefore both p1,p2p_1,p_2 can't be odd.

Therefore one of them must be an even prime p1=2\Rightarrow p_1=2 (Assuming p2>p1p_2>p_1). And if so 2=p3p2p12=p_3-p_2 \Rightarrow p_1 and p2p_2 are twin primes.

And according to the twin prime conjecture (not proved yet) there are infinitely many such primes.

Case 4 n=0n=0

As you will be seeing for n=0n=0 p1n+p2n=p3np10+p20=p201+1=12=1p_1^n+p_2^n=p_3^n \Rightarrow p_1^0+p_2^0=p_2^0\Rightarrow 1+1=1\Rightarrow \boxed{2=1} which leads to contradiction.

\therefore no solutions exists for n=0n=0

Case 5 n<0n<0

Let k=n;kZ+-k=n;k \in Z^+

Let us assume there exists primes p1,p2,p3p1n+p2n=p3np_1,p_2,p_3 | p_1^n+p_2^n=p_3^n p1n+p2n=p3n1p1k+1p2k=1p3kp_1^n+p_2^n=p_3^n \Rightarrow \frac{1}{p_1^k}+\frac{1}{p_2^k}=\frac{1}{p_3^k} p1k+p2kp1kp2k=1p3k\frac{p_1^k+p_2^k}{p_1^kp_2^k}=\frac{1}{p_3^k} p1kp2k=p3k(p1k+p2k){p_1^kp_2^k}=p_3^k(p_1^k+p_2^k) p1k=p3k(p1k+p2k)p2kp2kp3k(p12+p22){p_1^k}=\frac{p_3^k(p_1^k+p_2^k)}{p_2^k} \Rightarrow p_2^k | p_3^k(p_1^2+p_2^2) Either:p2kp3kp2p3\Rightarrow Either :p_2^k | p_3^k \Rightarrow \boxed{p_2 | p_3} Or:p2k(p1k+p2k)p2kp1kp2p1Or : p_2^k | (p_1^k+p_2^k) \Rightarrow p_2^k | p_1^k \Rightarrow \boxed{p_2 | p_1} Both of the above statements leads us to conclude that p2p_2 will always divide either of p3,p1p_3,p_1

\therefore All of p1,p2,p3p_1,p_2,p_3 can't be prime


Note :

  • aba|b means 'aa divides bb' or you may translate it as b0(moda)b ≡ 0 (\mod a)

  • Special thanks to Yajat Shamji to raise these type of questions

Note by Zakir Husain
3 weeks ago

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@Yajat Shamji -See here

Zakir Husain - 3 weeks ago

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Thanks...

What about Euler's identity (fully elaborated) proof?

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Proof for euler's identity is given on the brilliant itself see here, I also have the same proof for the formula.

Zakir Husain - 3 weeks ago

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@Zakir Husain Nice. Then what about Vieta's formula? Proof and fully elaborated on...

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@A Former Brilliant Member It is also give here for quadratic as well as higher degree polynomials

Zakir Husain - 3 weeks ago

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@Zakir Husain Ok. Final idea:

Can any arcprime (inverse prime) make a infinite series that is equivalent to a rational, transcendental or irrational number? Full proof...

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@A Former Brilliant Member I don't know about it, but know something called Highly composite number which James Grime referred as 'Anti-prime here.

Zakir Husain - 3 weeks ago

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@Zakir Husain You want to try, or should I try...

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@Zakir Husain LaTeX is:

p1=p^{- 1} = 1p\frac{1}{p}

Can 1p\frac{1}{p} be used in:

n=01p\sum_{n=0}^{\infty} \frac{1}{p}

Such that it produces a rational, irrational or transcendental number?

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@A Former Brilliant Member The sum of reciprocals of all primes is equal to \infty see here it was proved by Leonhard Euler

Zakir Husain - 3 weeks ago

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@Zakir Husain Well, I'm out! But about this note, can you do a note that proves that there is primes p1,p2,p3p_1, p_2, p_3 that satisfies for any nn:

p1n+p2n=p3n+1p_1^n + p_2^n = p_3^{n + 1}

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@A Former Brilliant Member I will try

Zakir Husain - 3 weeks ago

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@Zakir Husain Clue: there is already one:

22+112=532^2 + 11^2 = 5^3

2,11,52, 11, 5 are primes

Was in Squarimes numerical proof for primes 20\leq 20

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@A Former Brilliant Member For n=2n=2 there are infinitely many see: p12+p32=p33p331(mod4)p31(mod4)p_1^2+p_3^2=p_3^3\Rightarrow p_3^3\equiv 1 (\mod 4)\Rightarrow p_3\equiv 1 (\mod 4)

The only condition is that p31(mod4)p_3 \equiv 1 (\mod4) or p3=2p_3=2. If p3=2p2=p1=2p_3=2\Rightarrow p_2=p_1=2

Zakir Husain - 3 weeks ago

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@Zakir Husain Nice. Write a note on it...

A Former Brilliant Member - 2 weeks, 6 days ago

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@Zakir Husain @Yajat Shamji and @Mahdi Raza - I'm already having a handful topics to be written as notes, but I'm not getting enough time for them. But thanks for suggestions

Zakir Husain - 3 weeks ago

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