Number Theory #6

We let ${ 5 }^{ p }+4{ p }^{ 4 }={ k }^{ 2 }$ , $k>0$

${ k }^{ 2 }-4{ p }^{ 4 }={ 5 }^{ p }$

$(k-2{ p }^{ 2 })(k+2{ p }^{ 2 })={ 5 }^{ p }$

$\quad$

Case 1: $(k-2{ p }^{ 2 })=1$

$k+2{ p }^{ 2 }={ 5 }^{ p }$

$4{ p }^{ 2 }+1={ 5 }^{ p }$

It can be proved (using induction) that there is no solution for $p>1$ , thus there is no prime solution

Case 2: $(k-2{ p }^{ 2 })>1$

$5|(k-2{ p }^{ 2 }), (k+2{ p }^{ 2 })$

$5|(k+2{ p }^{ 2 })-(k-2{ p }^{ 2 })=4{ p }^{ 2 }$

$5|{ p }^{ 2 }$

$5|p$

$p=\boxed { 5 }$

Let $5^p + 4p^4 = k^2$ .

Then,

$(k+4p^2)(k-4p^2)=5^p$ .

Now we can observe that both $k + 4p^2$ & $k - 4p^2$ must both be powers of five themselves, if they multiply to give a power of five.

Hence,

$k + 4p^2 = 5^x$ and $k - 4p^2 = 5^y$ , where $x + y = p$ .

These can both be rearranged to give:

$k = 5^x - 4p^2$ and $k = 5^y + 4p^2$ .

Equating them gives:

$5^x - 4p^2 = 5^y + 4p^2$

$5^x - 5^y = 8p^2$

EDIT:

Case 1: both $x$ and $y$ $> 0$ .

All multiples of 5 must end in a 0 or a 5. Because there will be no 2 in the prime factorisation of a power of 5, all powers of 5 must end in 5. Hence, the differences between two powers of 5 must always be a multiple of 10.

$10m = 8p^2$

$5m = 4p^2$

Now we can see that $p^2$ must be a multiple of 5, in order for the equation to be true. But since $p$ is prime, the only multiple of 5 it can be is 5 itself.

Case 2: either $x$ or $y$ is $0$ .

Then: $5^x =8p^2$

This clearly has no solutions.

Hence there is only one solution, where $p = \boxed{5}$ .

$Let\quad \\ \quad \quad \quad \quad \quad { 5 }^{ p }+4{ p }^{ 4 }={ k }^{ 2 }\\ \Longrightarrow \quad { 5 }^{ p }=(k-2{ p }^{ 2 })(k+2{ p }^{ 2 })\\ \quad \quad \quad \quad \quad \because \quad p>0\\ \Longrightarrow \quad 5|(k-2{ p }^{ 2 })\quad \& \quad 5|(k+2{ p }^{ 2 })\\ \quad \quad \quad \quad \quad \quad or\\ k-2{ p }^{ 2 }\equiv 0\quad mod\quad 5\quad \longrightarrow 1\\ k+2{ p }^{ 2 }\equiv 0\quad mod\quad 5\quad \longrightarrow 2\\ \\ Adding\quad the\quad 2\quad congruences\\ \quad \quad \quad 2k\equiv 0\quad mod\quad 5\\ \Longrightarrow \quad k\equiv 0\quad mod\quad 5\\ \\ Substituting\quad k\quad in\quad the\quad 2nd\quad Congruence\\ \quad \quad 2{ p }^{ 2 }\equiv 0\quad mod\quad 5\\ \Longrightarrow \quad p\equiv 0\quad mod\quad 5\quad \quad (\because \quad 5\quad is\quad prime\quad \& \quad 5|{ p }^{ 2 }\quad and\quad so\quad 5|{ p })\\ \\ \because \quad p\quad is\quad prime\quad and\quad 5|{ p }\\ \quad \quad p=5\quad is\quad the\quad only\quad solution.$

I am not sure if this is correct but I used modular arithmetic. For primes 2,3,7 and above 7 the we get 4= n^2 mod 5 except for 5 which there is no remainder. So only p =5 works. Please correct me if my method is wrong.