$n$ is natural number and $p$ is prime number. If $1+np$ is square of natural number then prove that $n+1$ is equal to some sum of $p$ square of natural numbers.
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Tags: number theory, prime numbers
Sardor
05.06.2016 08:39
It's from Poland MO.
roopsa
06.06.2016 15:40
What is the meaning of"sum of p squares of natural numbers".
dominicleejun
18.11.2016 20:14
http://www.artofproblemsolving.com/community/c6h369941 http://www.artofproblemsolving.com/community/c6h569731 does anybody know which yr Poland MO this qn is from?
hectorleo123
16.05.2023 03:54
shohvanilu wrote: $n$ is natural number and $p$ is prime number. If $1+np$ is square of natural number then prove that $n+1$ is equal to some sum of $p$ square of natural numbers. It is also Peru 2003 Cono Sur TST P2 https://artofproblemsolving.com/community/c6h3065494p27657509
hectorleo123
16.05.2023 03:59
shohvanilu wrote: $n$ is natural number and $p$ is prime number. If $1+np$ is square of natural number then prove that $n+1$ is equal to some sum of $p$ square of natural numbers. We have $1+np=a^2 \Rightarrow a \equiv 1 \pmod{p}$ or $a\equiv -1 \pmod{p}$ If $a \equiv 1 \pmod{p}$: $\Rightarrow a=px+1 \Rightarrow n+1=(p-1)x^2+(x+1)^2_\blacksquare$ If $a \equiv -1 \pmod{p}$: $\Rightarrow a=px-1 \Rightarrow n+1=(p-1)x^2+(x-1)^2_\blacksquare$
RenheMiResembleRice
16.05.2023 06:11
hectorleo123 wrote: shohvanilu wrote: $n$ is natural number and $p$ is prime number. If $1+np$ is square of natural number then prove that $n+1$ is equal to some sum of $p$ square of natural numbers. We have $1+np=a^2 \Rightarrow a \equiv 1 \pmod{p}$ or $a\equiv -1 \pmod{p}$ If $a \equiv 1 \pmod{p}$: $\Rightarrow a=px+1 \Rightarrow n+1=(p-1)x^2+(x+1)^2_\blacksquare$ If $a \equiv -1 \pmod{p}$: $\Rightarrow a=px-1 \Rightarrow n+1=(p-1)x^2+(x-1)^2_\blacksquare$ Why does the RHS follow in $n+1=(p-1)x^2+(x+1)^2_\blacksquare$
hectorleo123
17.05.2023 00:17
RenheMiResembleRice wrote: hectorleo123 wrote: shohvanilu wrote: $n$ is natural number and $p$ is prime number. If $1+np$ is square of natural number then prove that $n+1$ is equal to some sum of $p$ square of natural numbers. We have $1+np=a^2 \Rightarrow a \equiv 1 \pmod{p}$ or $a\equiv -1 \pmod{p}$ If $a \equiv 1 \pmod{p}$: $\Rightarrow a=px+1 \Rightarrow n+1=(p-1)x^2+(x+1)^2_\blacksquare$ If $a \equiv -1 \pmod{p}$: $\Rightarrow a=px-1 \Rightarrow n+1=(p-1)x^2+(x-1)^2_\blacksquare$ Why does the RHS follow in $n+1=(p-1)x^2+(x+1)^2_\blacksquare$ I do not understand your question