Question

if n is an integer and 2+2✓(28n2+1) is an integer.....prove that it is a perfect integer?

Answer 1

Hello,

Notice that 2+228n2+1−−−−−−−√2+228n2+1 is an even integer. Also, 28n2+128n2+1 is a perfect square of an odd integer say mm (Because 28n2+128n2+1 is odd itself).

Now,

28n2=m2−1=(m+1)(m−1)⟹7n2=(m+12)(m−12)28n2=m2−1=(m+1)(m−1)⟹7n2=(m+12)(m−12)

Hence, (m+12)=7a2(m+12)=7a2,(m−12)=b2(m−12)=b2 or (m+12)=b2(m+12)=b2, (m−12)=7a2(m−12)=7a2. This is because 7n27n2 is 77 times of a square and thus the right side is also 77 times a perfect square. This is only possible when one of them is 77 times of a square and other is simply a square as (Square∗Square=Square)(Square∗Square=Square), you can say that there is possibility that they both are not squares but the product is (like 2∗8=16=422∗8=16=42), but notice that (m+12)(m+12) and (m−12)(m−12) are consecutive integers and hence coprime

If m+12=7a2m+12=7a2 and m−12=b2m−12=b2 then b2≡−1mod(7)b2≡−1mod(7), a contradiction.

Hence, m−12=7a2m−12=7a2 and m+12=b2m+12=b2. Hence, 2+2m=4b22+2m=4b2 a perfect square.

Hope i  helped u 

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