## Jonask 2 years ago For how many positive integer values of c does the equation below have an integer solution?

\[\huge \color{brown}{ 2x^2+689x+c}\]

@TuringTest @phi @hartnn @sauravshakya @shubhamsrg

\[(2x+a)(x+b)=4x^2+2x(a+b)+ab\] \[2(a+b)=689,ab=c\]

i dont kow wat to do here

5. TuringTest

@JamesJ D:

first 689 can never be written as even number

7. TuringTest

I am pretty bad at things like this for some reason, sorry.

thanks Turing

wolfram gives interger solutions as \[c=689n-2n^2, x=-n\]

11. TuringTest

I am assuming the equation is\[\large2x^2+689x+c=0\]right?

13. shubhamsrg

2x^2 + 689x + c = 0 c = -689x - 2x^2 x= n will always given a -ve c, hence x should be -n , for some integer n, => c = 689n - 2n^2 this has to be >0 => 689n -2n^2 >0 => n(689 - 2n) >0\ We already know n>0 hence we only have to deal with 689 > 2n or n < 344.5 Hence there can be 344 required value's of c. I have done this on brilliant.org if I am not wrong.

yes its from them,i have many other problems from them,if you dont mind you can check the question i asked bfore this one...thanks makes sense

is this the only c

16. shubhamsrg

What do you mean "the only c" ?

the question says For how many positive integer values of c so are there 344 interger c's

18. shubhamsrg

yes.