What name shall I give to this problem? - 3

Number Theory Level 3

Let S S be the sum of all integers n n such that

3 n − 5 n + 1 {\sqrt{\dfrac{3n-5}{n+1}}}

is an integer. Find the value of S 2 S^{2} .

This problem is from the set What name should I give? .


The answer is 36.

View wiki
Surya Prakash by Surya Prakash , 22, India

This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try refreshing the page, (b) enabling javascript if it is disabled on your browser and, finally, (c) loading the non-javascript version of this page . We're sorry about the hassle.

1 solution

Curtis Clement
Aug 26, 2015

3 n − 5 n + 1 = 3 ( n + 1 ) − 8 n + 1 = 3 − 8 n + 1 \frac{3n-5}{n+1} = \frac{3(n+1) -8}{n+1} = 3-\frac{8}{n+1} Now this has to be a square so we have n= -9, 3. ∴ S 2 = ( 9 − 3 ) 2 = 6 2 = 36 \therefore\ S^2 = (9-3)^2 = 6^2 = \boxed{36}

4 Helpful 0 Interesting 0 Brilliant 0 Confused

Can explain how you prove that for 3 − 8 n + 1 3-\frac{8}{n+1} to be square n must be -9 or 3

I know that -9 and -3 are the only options - but I proved it another way - how did you prove it.

Tony Flury - 5 years, 9 months ago

Log in to reply

I looked at the factors of 8 (such that 8 n + 1 ≤ 3 \frac{8}{n+1} \leq\ 3 ) and hence tested all the values of 3 − 8 n + 1 \ 3 - \frac{8}{n+1}

Curtis Clement - 5 years, 9 months ago

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...