Question

Determinati numerele ab aftfel incat B= ab+5ba-126 sa fie patrat perfect.

Aratati ca numerele de forma $10^{2n}$ + 5 nu pot fi patrate perfecte 

Answer 1

$ab+5ba-126 \geq 0 \\ab+5ba \geq 126 \\ B=ab+5ba-126=10a+b+50b+5a)-126=15a+51b-126= \\ =3(5a+17b-42) \\ 5a+17b-42=3 ^{putere.impara} \\ 5a+17b-42=divizibil.cu.3 \\ 42=divizibil.cu.3 \\ 5a+17b=divizibil.cu.3 \\ a=divizibil.cu.3\\b=divizibil.cu.3 \\ a=3;6;9 \\ b=3;6;9 \\ a=3;b=6;ab+5ba-126= \\ =36+5*63-126=225= 15^{2} \\ a=3;b=9;ab+5ba-126= \\ =39+5*93-126=378=nu.este.pp \\ a=6;b=3;ab+5ba-126= \\=63+5*36-126=117=nu.este.pp \\ a=6;b=9;ab+5ba-126= \\=69+5*96-126=423=nu.este.pp$
$a=9;b=3;ab+5ba-126= \\=93+5*39-126=162=nu.este.pp \\ a=9;b=6;ab+5ba-126= \\=96+5*69-126=315=nu.este.pp$

$10^{2n+5}= 10^{2(n+2)+1}= 10 ^{2(n+2)}*10= (10 ^{n+2} ) ^{2}*10$