Even I don't know the answer, that's very odd actually.

Number Theory Level 2

$a; \ b; \ x; \ y$ are integers that satisfy: $\large x = a^2b + b^2a + ab + a + b$ $\large y = a^3b^2 + b^3a^2 + 2a^2b^2 + a^3b + b^3a$ If there are an odd number and an even number in $x$ and $y$ , which is which?

x

is even,

x

is odd.

x

is odd,

y

is even. Cannot be determined.

2 solutions

Pham Khanh
Aug 20, 2017

$y=a^3b^2+b^3a^2+2a^2b^2+a^3b+b^3a$ $=(a^3b^2+a^3b)+(b^3a^2+b^3a)+2a^2b^2$ $=a^3{\color{#D61F06}{b(b+1)}} +b^3{\color{#20A900}{a(a+1)}}+2a^2b^2$ As $\color{#D61F06}{b;b+1}$ are 2 consecutive integers, there must be an even number, therefore $\color{#D61F06}{b(b+1)}$ is divisible by 2. Also, we have $\color{#20A900}{a(a+1)}$ is divisible by 2. So y is even .But, for example, $a=b=1$ , $x=5$ is odd. So if there is an odd integer, it must be $x$ . so $\boxed{x~i s~o d d,~y~i s~e v e n}$

Aaghaz Mahajan
Apr 11, 2019

@Thành Đạt Lê Well, $x$ can't be both odd and even at the same time........

Even I don't know the answer, that's very odd actually.

2 solutions

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