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Find positive integer numbers $a,b,c$ such that all the roots of these quadratic polynomials are positive integer numbers

1) $x^2−2ax+b=0$

2) $x^2−2bx+c=0$

3) $x^2−2cx+a=0$

Sorry but I don't know what to do :(( please help me :((

2 Answers 2

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The only solution in positive integers is $a=b=c=1$.

For the equations to have integer roots, their discriminants must be perfect squares. Taking the first equation for example, this means $a^2-b$ must be a perfect square. But $a^2-b \lt a^2$ thus it must be less than or equal the next smaller perfect square below $a^2$ which is $(a-1)^2$. Therefore:

$$a^2-b \le (a-1)^2 = a^2 - 2a +1 \quad \iff \quad b + 1 \ge 2a$$

Adding the $3$ similar inequalities together gives:

$$a+b+c+3 \ge 2(a+b+c) \quad \iff \quad a+b+c \le 3$$

The latter has $a=b=c=1$ as the unique solution in positive integers, in which case all $3$ equations reduce to $x^2-2x+1=(x-1)^2=0$ with the positive integer $1$ as a double root.

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    $0=0$ does not constrain anything. However chaining the inequalities gives $a^2\ge b^2\ge c^2\ge a^2$ which means all these are identical. – 2017-01-07
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    @Macavity If $u \ge 0$, $v \ge 0$ and $u+v=0$ then $u=v=0$. – 2017-01-07
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    I thought that the discriminant is $ a^2-b $ :/ – 2017-01-07
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    Actually my question is why $ a^2-b^2 > 0 $ – 2017-01-07
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    @UchihaItachi You are right, thanks for pointing. I reworked my answer. – 2017-01-07
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    wow :o it's incredible >>.<< – 2017-01-08
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Hint:

The trivial case where $a=b=c=1$ gives you the monic polynomial $x^2 - 2x + 1$ which by Descarte's Rule of signs has two positive roots and by rational root theorem has two roots both $x=1$.