Question

Show that x²+7x-14(q²+1)=0 where q € I has no integral roots​

Answer 1

Answer:

Suppose k∈Z is an integer solution to this equation. Then, we have

k(k2+7)=2×7×(n2+1).

Since 7 is a factor on the RHS, we must have either 7|k or 7|(K2+7) , i.e., 7|k2 . In either case, 7|k. So, taking k=7m , we get

7m(49m2+7)=2×7×(n2+1), i.e.,7m(7m2+1)=2×(n2+1).

Now, since 7|LHS and 7/|2 , we must have 7|n2+1 . Let n=7r+l where l=0,1,2,3,4,5,6 and so that l2+1=1,2,5,10,17,26,37 . If 7|(7r+l)2+1 , we must have 7|l2+1. But clearly 7/|l2+1 . So, we have reached a contradiction.

Step-by-step explanation:

DANGER JATTNI