What are the solutions if h+k=2016 and the roots of x^2+hx+k=0 are integers?

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SUMMARY

The equation \( h + k = 2016 \) is established as a constraint for the quadratic equation \( x^2 + hx + k = 0 \) to have integer roots. By applying Vieta's formulas, it is concluded that the roots can be expressed as \( r_1 \) and \( r_2 \), where \( r_1 + r_2 = -h \) and \( r_1 r_2 = k \). The integer nature of the roots leads to the derivation of specific pairs \( (h, k) \) that satisfy both conditions, resulting in multiple valid solutions such as \( (2016, 0) \) and \( (0, 2016) \).

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Albert1
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$if$ $ h+k=2016$
and all roots of $ x^2+hx+k=0$ are both integers
find its solutions
 
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Albert said:
$if$ $ h+k=2016$
and all roots of $ x^2+hx+k=0$ are both integers
find its solutions

let the roots be - a and - b and |a| <= |b| so we get
$(x+a)(x+b) = x^2+hx + k$
or $x^2 + (a+b) x + ab = x^2 + hx + k$
so $a+ b = h,ab = k$
hence $(1+a)(1+b) = (1+ a + b + ab) = 1 + h + k = 2017$
it is a prime so 1 + a = 1 and 1 + b = 2017 so a = 0 and b = 2016 and roots are 0 and - 2016. h = 2016 and k = 0 one solution
or 1 + a = - 1 and 1+b = - 2017 giving a = -2 and b = - 2018 another solution so 2 solutions
 
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