2

This is the original question that was posted and answered. For the question the OP intended to ask, please see the edit history.

$a,b$ are positive integers. Prove that if $k=\frac{a^2+b^2}{a+b+1}$ is an integer then $k=5$.

I tried to see the whole expression as a quadratic of $a$ but that is not helping much.

Caleb Stanford
  • 45,895
  • What do you get if you see the whole expression as a quadratic in $a$? – Kenny Lau Apr 28 '16 at 15:51
  • What is the background for this.. It would be of some help if you can say that.. Some body who wants to try can guess where to start... –  Apr 28 '16 at 16:10
  • $$k=\frac{a^2+b^2}{a+b+1}$$ $$a=2p^2+p$$ $$b=2p^2-p$$ $$k=2p^2$$ – individ Apr 28 '16 at 16:34
  • http://www.artofproblemsolving.com/community/c3046h1056472_one_binary_form – individ Apr 30 '16 at 06:13
  • Ayan, can you ask the new question by clicking "Ask Question" at the top of the page? It seems like an interesting question. Unfortunately, you shouldn't edit this post because this was a different question that was already asked and answered by several people. – Caleb Stanford Sep 16 '16 at 15:20

2 Answers2

2

Nonsense. $a=105, b=7, k=98$. How did you get that idea? Or $a=3, b=1, k=2$.

1

In this case, if $k=5,$ then we would have $70$ as the sum of two integer squares, which it is not.

$$ a^2 + b^2 = 5a + 5b + 5, $$ $$ 4a^2 + 4 b^2 = 20a + 20 b + 20, $$ $$ 4 a^2 - 20a + 4 b^2 - 20 b = 20, $$ $$ 4 a^2 - 20a + 25 + 4 b^2 - 20 b + 25 = 20 + 25 + 25 = 70, $$ $$ (2a-5)^2 + (2b-5)^2 = 70. $$ However, $70 = 2 \cdot 5 \cdot 7$ is not the sum of two integer squares.

A superficially similar problem that does lead to $5$ is $$ \frac{a^2 + b^2}{ab-1} = k, $$ see Is it true that $f(x,y)=\frac{x^2+y^2}{xy-t}$ has only finitely many distinct positive integer values with $x$, $y$ positive integers? and my three or so answers there. I did, with help, eventually settle the whole thing.

If the version I have guessed is really the original problem, then it fits under https://en.wikipedia.org/wiki/Vieta_jumping

Community
  • 1
Will Jagy
  • 139,541
  • Actually there is a slightly different task. $$k=\frac{a^2+b^2}{a+b+1}$$ Write in such a formal way - so better. $$(2a-k)^2+(2b-k)^2=2k(k+2)$$ – individ Apr 28 '16 at 16:49