Problem to Brilliant!
Find the sum of all positive integers such that each of the equations has distinct positive integral roots.
The answer is 11.
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
Let integers α>β>0 be the roots of (i) x² -ax+ b=0 and let integers γ>δ>0 be roots of (ii) x² - bx + a= 0 . For definiteness, let a ≥ b. Now α + β= a , αβ=b , and γ + δ=b , γδ =a Hence a-b =1-(α -1) (β-1) . Hence 0 ≤ 1 – (α -1) (β-1) ≤ 1. So β=1 since α , β are positive integers and β< α .Thus a-b =1. Further, a- b =(γ-1) (δ -1)-1 , so that (γ-1) (δ-1)=2. So since γ>δ>0 are integers we see that γ -1=2 and δ -1 =1 ,so that γ =3 ,δ=2. Hence , a= γδ =6 and b= γ+δ =5 .Also ,therefore, α=5, β= 1 and a= α + β=5+1=6 , b= αβ=5 therefore a+b=11