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Fara a efectua calculele, stabiliti daca numarul natural

\( x = 2001\cdot2002\cdot2003\cdot2004 + 2004\cdot2005\cdot2006\cdot2007-(2004\cdot2007 + 1)\cdot(2004\cdot4002+2)+2 \)

este patrat perfect.

Notam n=2001 si avem
x=n(n+1)(n+2)(n+3)+(n+3)(n+4)(n+5)(n+6)-{[(n+3)(n+6)+1][2n(n+3)+2]+2

x=n(n+1)(n+2)(n+3)+(n+3)(n+4)(n+5)(n+6)-2n(n+3)(n+3)(n+6)-2n(n+3)-
-(n+3)(n+6)-2+2

x=n(n+1)(n+2)(n+3)-2n(n+3)+(n+3)(n+4)(n+5)(n+6)-2n(n+3)(n+3)(n+6)-
-2(n+3)(n+6)

x=n(n+3)[(n+1)(n+2)-2]+(n+3)(n+6)[(n+4)(n+5)-2n(n+3)-2]

x=n(n+3)(n^2+3n+2-2)+(n+3)(n+6)(n^2+9n+20-2n^2-6n-2)

x=n^2*(n+3)^2+(n+3)(n+6)(-n^2+3n+18 )

x=n^2*(n+3)^2-(n+3)(n+6)(n+3)(n-6)

x=(n+3)^2*[n^2-(n^2-36)]

x=(n+3)^2*36
x=[6*(n+3)]^2
revenind x=(6*2004)^2

adica x=(12024)^2

Felicitari! Corect!

Mie nu-mi prea plac calculele lungi, prin urmare am calculat toate patratele perfecte, cuprinse intre 100000000 si 1000000000, cu creionul, in cateva ore, (8$)/ora = >\( 8\cdot(8$) = 64$ \), printr-o metoda... . Am gasit 2122 de patrate perfecte cu ultima cifra 5.
Asta a fost OFF ... Daca nu glumesc, mi se blocheaza intelighenchia, dar am spus drept

Bun!

Mai este o cale de rezolvare, mult mai simplu, mai putine calcule.

Natalee

produs de 4 numere consecutive = diferenta de patrate
(n+1)(n+2)(n+3)(n+4) = [(n+1)(n+4) +1]^2 - 1 etc.