Is it an integer?

Number Theory Level 3

Let $a,b, \text{ and } c$ be positive integers such that $\large\frac{a\sqrt{2017}+b}{b\sqrt{2017}+c}$ is a rational number.

Is $\large\frac{a^2+b^2+c^2}{a+b+c}$ an integer?

Yes No

2 solutions

Tom Engelsman
Feb 9, 2018

If $\frac{a\sqrt{2017} + b}{b\sqrt{2017} + c} = \frac{p}{q}$ (where $a, b, c, p, q \in \mathbb{N}$ ) be a rational number, then one obtains:

$(aq - bp)\sqrt{2017} = cp - bq$ (i)

But $2017$ is a prime $\Rightarrow \sqrt{2017} =$ irrational. This forces both sides of (i) to be equal to zero, which occurs iff $a = b = c$ AND $p = q$ . Hence, the quantity:

$\frac{a^2 + b^2 + c^2}{a+b+c} = \frac{3a^2}{3a} = a$

is an integer.

$\mathbb{Q.E.D.}$

Nice solution as usual.

Hana Wehbi - 3 years, 4 months ago

Still beatin' last year's "2017-is-prime" horse, Hana! Good prob, thanks :)

tom engelsman - 3 years, 4 months ago

I’m glad you like my problems, at least, there is something positive about them.

Hana Wehbi - 3 years, 4 months ago

The solution is slightly incorrect. The statement "This forces both sides of (i) to be equal to zero, which occurs iff $a=b=c$ AND $p=q$ " is false. "The correct statement is "This forces both sides of (i) to be equal to zero, which occurs iff $a / b = b / c = p / q$ . Then, we can show that ${a^{2} + b^{2} + c^{2}} / {a + b + c} = a - b + c$ OR $c - b + a$ , depending on whether $a \geq c$ or $c \geq a$ .

Krutarth Patel - 5 months, 3 weeks ago

Rab Gani
Aug 17, 2019

Similar with Tom's solution. with aq-bp = 0, and cp-bq=0. then we have a/b = b/c = p/q. So we have b^2 = a.c. So the solutions a,b,c = a,2a,4a. If we substitute to (a^2+b^2+C^2)/(a+b+c) = (21a^2)/(7a) = 3a which is integer.

Is it an integer?

2 solutions

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