# Power sum of primes

In this discussion I will prove the following statement:

The equation $p_1^n+p_2^n=p_3^n$ will have solutions where $n \in \mathbb Z$ and all of $p_1,p_2,p_3$ are primes if and only if $n=1$.

Case 1 $n>2$

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

Case 2 $n=2$

Let us assume there exists primes $p_1,p_2,p_3 | p_1^2+p_2^2=p_3^2$ $p_1^2=p_3^2-p_2^2=(p_3-p_2)(p_3+p_2)$ Now pairwise factors of $p_1^2$ are $(p_1^2,1),(p_1,p_1)$ now if $p_3-p_2=p_1\Rightarrow p_3-p_2=p_1=p_3+p_2\Rightarrow p_2=0$ which is not possible $\therefore p_3-p_2=1;p_3+p_2=p_1^2 \because p_3-p_2 But this also leads to contradiction because according to it $p_3=p_2+1$ which is not possible for $p_3>3$ ($\because$ if $p_3$ is odd then $p_2$ will be divisible by $2$ and vice versa)

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

Therefore no solution exists for $n=2$

Case 3 $n=1$

Let us assume there exists primes $p_1,p_2,p_3 | p_1+p_2=p_3$. $p_1+p_2=p_3$ In the above equation, if all of $p_1,p_2,p_3$ are primes, then $p_3>p_2;p_3>p_1$ therefore $p_3$ is greater than the $2_{nd}$ prime $\Rightarrow p_3>3$.

$\Rightarrow p_3$ is odd $\therefore$ both $p_1,p_2$ can't be odd.

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

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

Case 4 $n=0$

As you will be seeing for $n=0$ $p_1^n+p_2^n=p_3^n \Rightarrow p_1^0+p_2^0=p_2^0\Rightarrow 1+1=1\Rightarrow \boxed{2=1}$ which leads to contradiction.

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

Case 5 $n<0$

Let $-k=n;k \in\mathbb Z^+$

Let us assume there exists primes $p_1,p_2,p_3 | p_1^n+p_2^n=p_3^n$ $p_1^n+p_2^n=p_3^n \Rightarrow \frac{1}{p_1^k}+\frac{1}{p_2^k}=\frac{1}{p_3^k}$ $\frac{p_1^k+p_2^k}{p_1^kp_2^k}=\frac{1}{p_3^k}$ ${p_1^kp_2^k}=p_3^k(p_1^k+p_2^k)$ ${p_1^k}=\frac{p_3^k(p_1^k+p_2^k)}{p_2^k} \Rightarrow p_2^k | p_3^k(p_1^2+p_2^2)$ $\Rightarrow Either :p_2^k | p_3^k \Rightarrow \boxed{p_2 | p_3}$ $Or : p_2^k | (p_1^k+p_2^k) \Rightarrow p_2^k | p_1^k \Rightarrow \boxed{p_2 | p_1}$ Both of the above statements leads us to conclude that $p_2$ will always divide either of $p_3,p_1$

$\therefore$ All of $p_1,p_2,p_3$ can't be prime

Note :

• $a|b$ means '$a$ divides $b$' or you may translate it as $b ≡ 0 (\mod a)$

• Special thanks to Yajat Shamji to raise these type of questions Note by Zakir Husain
1 year ago

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## Comments

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@Yajat Shamji -See here

- 1 year 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.

- 1 year ago

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

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It is also give here for quadratic as well as higher degree polynomials

- 1 year ago

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

- 1 year ago

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You want to try, or should I try...

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LaTeX is:

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

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

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

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

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The sum of reciprocals of all primes is equal to $\infty$ see here it was proved by Leonhard Euler

- 1 year ago

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Well, I'm out! But about this note, can you do a note that proves that there is primes $p_1, p_2, p_3$ that satisfies for any $n$:

$p_1^n + p_2^n = p_3^{n + 1}$

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I will try

- 1 year ago

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Clue: there is already one:

$2^2 + 11^2 = 5^3$

$2, 11, 5$ are primes

Was in Squarimes numerical proof for primes $\leq 20$

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For $n=2$ there are infinitely many see: $p_1^2+p_3^2=p_3^3\Rightarrow p_3^3\equiv 1 (\mod 4)\Rightarrow p_3\equiv 1 (\mod 4)$

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

- 1 year ago

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Nice. Write a note on it...

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@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

- 1 year ago

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