$ax+by$

Number Theory Level 3

Let $a$ and $b$ be two distinct positive integers.

Is it always possible to find two positive integers $x$ and $y$ such that

$\binom{x+y}{2}=ax+by?$

Yes, it is always possible No, it is only possible when

a+b

is odd No, it is never possible.

1 solution

Levi Walker
Oct 28, 2018

We can expand the binomial coefficient as $\binom{x+y}{2} = \frac{(x+y)!}{2(x+y-2)!} = \frac{1}{2}(x+y)(x+y-1) = \frac{1}{2}x(x + 2y -1) + \frac{1}{2}y(y -1)$ Setting $2a=(x + 2y -1)$ and $2b= (y -1)$ turns the expression into $ax + by$ as required.

Edit: This solution does not guarantee that x and y will be positive.

What guarantees that the system you provided will give $x,y$ that satisfy the requirements of problem (positive integers). Put in $b = 4, a = 2$ and your system gives $y=9, x=-13$ .

Leonel Castillo - 2 years, 7 months ago

I totally forgot about that condition. I'm not sure if such a system is possible, then.

Levi Walker - 2 years, 7 months ago

I found a more general way of finding solutions and I have already found cases (again, take b=4, a=2) for which no solution exists. I suspect that the solution could be that only solutions exist when a+b is odd (as so far my method has found solutions for all instances of that case I tested), however, I have not continued to study the problem since I found the one who posted the problem was wrong. Who knows if any of the options given are valid now. I'll wait for an update.

Leonel Castillo - 2 years, 7 months ago

@Leonel Castillo – If $a=2, b=4$ , them $x=2, y=6$ satisfies the conditions.

Áron Bán-Szabó - 2 years, 7 months ago

@Áron Bán-Szabó – It seems I have also made a mistake. I'll revisit to see where I went wrong.

Leonel Castillo - 2 years, 7 months ago

a x + b y ax+by a x + b y

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$ax+by$