find number of integers n such that n+16/n+6 is an integer
Answers
Answer:
you can take some help for the given answers
Step-by-step explanation:
Let n+9=a^2,16n+9=b^2 and 27n+9=c^2.
a^2-9=(b^2-9)/16=(c^2-9)/27
=>16a^2-b^2=135
=>(4a+b)(4a-b)=135=15*9=27*5=45*3
=135*1
=>4a+b=15,4a-b=9=>(a,b)=(3,3) not possible.
4a+b=27,4a-b=5=>(a,b)=(4,11)=>n=7
4a+b=45,4a-b=3=>(a,b)=(6,21)=>n=27
4a+b=135,4a-b=1=>(a,b)=(17,67)=>
n=280
n=7=>27n+9=198not a perfect square
n=27=>27n+9=738not a perfect square
n=280=>27n+9=7569=(87)^2
Hence n=280 is only possible solution.
Hope this works......
Answer:
Let n+9=a^2,16n+9=b^2 and 27n+9=c^2.
a^2-9=(b^2-9)/16=(c^2-9)/27
=>16a^2-b^2=135
=>(4a+b)(4a-b)=135=15*9=27*5=45*3
=135*1
=>4a+b=15,4a-b=9=>(a,b)=(3,3) not possible.
4a+b=27,4a-b=5=>(a,b)=(4,11)=>n=7
4a+b=45,4a-b=3=>(a,b)=(6,21)=>n=27
4a+b=135,4a-b=1=>(a,b)=(17,67)=>
n=280
n=7=>27n+9=198not a perfect square
n=27=>27n+9=738not a perfect square
n=280=>27n+9=7569=(87)^2
Hence n=280 is only possible solution.
Hope this works......
Step-by-step explanation:
hope it would be helpful