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For how many positive integer values of c does the equation below have an integer solution?

Algebra
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\[\huge \color{brown}{ 2x^2+689x+c}\]
\[(2x+a)(x+b)=4x^2+2x(a+b)+ab\] \[2(a+b)=689,ab=c\]

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Other answers:

i dont kow wat to do here
first 689 can never be written as even number
I am pretty bad at things like this for some reason, sorry.
thanks Turing
http://www.wolframalpha.com/input/?i=689
wolfram gives interger solutions as \[c=689n-2n^2, x=-n\]
I am assuming the equation is\[\large2x^2+689x+c=0\]right?
http://www.wolframalpha.com/input/?i=2x%5E2%2B689x%2Bc%3D0
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
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
yes.
okay thanks

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