Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Does anyone know how to find integer solutions of the quadratic equation


where $z$ is a fixed odd prime or $1$ and $f$ is a fixed odd prime greater than $3$?

This problem arose from the Diophantine equation $A+B=C$ where $A,B,C$ are natural numbers with no common factor. The managers of this site asked me to make my questions harder for this reason I will restate the above. Does anyone know if the quadratic equation $x^2-2x-[a^5+b^5]=0$ has infinite integer solutions?

share|cite|improve this question
The quantification is not clear. Are $z$ and $f$ previously-given parameters, or what? – Lubin Sep 29 '12 at 19:08
@Lubin.only one parameter is given. In sort the question asks to find the pair of primes such as that.f-z=y[y+1].We know the difference of two primes can be expressed as the difference of two squares.So the above is saying there are infinite pairs of primes which their difference is expressed as above. – Vassilis Parassidis Sep 29 '12 at 19:20
The difference of two primes always has two factors which are the following.2[2x-2y+1] or 4[x-y] which are obtained from the difference of their squares. – Vassilis Parassidis Sep 29 '12 at 22:23
"We know the difference of two primes can be expressed as the difference of two squares." 11 and 5 are prime, and their difference is 6. How do you propose to express 6 as a difference of two squares? – Gerry Myerson Oct 2 '12 at 0:43

$$ y^2 + y + z = f $$ $$ y^2 + y + z - f = 0 $$ $$ ay^2 + by + c = 0 $$ where $a=1$, $b=1$, and $c=z-f$. So $$ y = \frac{-b\pm\sqrt{b^2-4ac}}{2a} = \frac{-1\pm\sqrt{1-4(z-f)}}{2} $$ In its method of solution, it's no different from any other quadratic equation.

Whether the solutions are integers depends of course on $z$ and $f$.

share|cite|improve this answer
My guess is that what OP is asking is, given a prime $f$, how do you find $y$ and prime $z$ such that $y^2+y+z=f$. – Gerry Myerson Oct 2 '12 at 6:34
If that's what he meant, he certainly wasn't clear about it. – Michael Hardy Oct 2 '12 at 16:01

Why dont you express the equation in the form $y^2+y+(z-f)=0$ and use the discriminant $b^2-4ac$ where $a=1$, $b=1$ and $c= z-f$. $c$ could be the difference of two primes or check the sequence. find the possible values of $y$.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.