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Let $t$ be a real number such that $t^2 = at + b$ for some positive integers $a$ and $b$. Then for any choice of positive integers $a$ and $b$, $t^3$ is not equal to -

1) $4t+3$

2) $8t+5$

3) $10t+3$

4) $6t+5$

This question is from a prestigious Indian Scholarship exam, and I'm solving it for fun. But I can't get at the bottom of the problem. I'm also a bit shaky on which concept to use in this problem.

My try on this question

$t^2 =at +b$
$t = \sqrt(at+b)$

For $t \in R$
$at+b \ge 0$

Since $a$ and $b$ are positive integers $at \ge -b$ and

$t \ge \frac{-b}{a}$

I also tried differentiating the function $t^2 = at + b$ to get

$\frac{d}{dt}(t^2) = a\frac{d}{dt}(t) + 0$

$2t = a$

Alas it leads nowhere. But

$\frac{a}{2} \ge \frac{-b}{a}$

Since $a$ and $2$ are positive integers

$a^2 \ge -2b$

What has to done next? Please help

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    Your very last step is trivial as we're given $\;a,b>0\;$ ... – 2017-01-14
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    Please suggest a right step @DonAntonio – 2017-01-14
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    @Y Thinking now... – 2017-01-14

2 Answers 2

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Hint:$$t^3 = t^2\cdot t \\= (at+b)\cdot t \\= at^2 + bt \\= a(at+b) + bt \\= (a^2 + b)t + ab$$

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    What should be the next step? – 2017-01-14
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    I am quite certain that you know that already, @labbhattacharjee. I am not inclined to give more until the OP has read this and had a go at cracking it themself. That's why I put "Hint" at the top. – 2017-01-14
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    By plugging from the list of options we can reach to the correct answer.. – 2017-01-14
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    @YourAverageEuler This does lend itself to a brute force method, yes. Mostly because you don't have too many possibilities to check. If the options were more along the lines of $1934t + 782$, you would have to be a bit more clever. It also helps that all the constant terms in all the options are primes. – 2017-01-14
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    In the most general setting, I think it'd be enough to consider stuff modulo $\;8\;$ to solve . – 2017-01-14
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    @Arthur since $a,b$ are positive integers... from your solution we can assume $a=1$ and $b=3$ and vice versa. That gives us the 1st and 3rd option. For $a=5$ and $b=1$ and we get $26t + 5$ and by switching $a$ and $b$ we get the 4th option. This means 2nd option is not possible. – 2017-01-14
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    Thanks @Arthur Is there a better way to do it, like a general case – 2017-01-14
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    @YourAverageEuler Nice, and as Arthur said: checking all the possibilities the second one is ruled out as for the other three we can get example. The interesting thing would be if you had more options to choose from... – 2017-01-14
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    @YourAverageEuler That's the solution, yes. Note that we can prove that the 2nd option is impossible (not just "difficult" or "we couldn't find a solution", but actually impossible) becase we must have $ab = 5$, which forces $a = 1, b = 5$ or vice versa, and neither gives $a^2 + b = 8$. – 2017-01-14
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    @DonAntonio yes you are right. But even this was tough for me and I was on the wrong track to solve this one. God knows when I'll be able to think such methodically – 2017-01-14
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    @YourAverageEuler There isn't so much a method to problem solving as just being able to see what approaches are easy and fast, and _might_ give a solution, so you just check those before you go on to the heavier stuff. – 2017-01-14
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    @Arthur nice point. Thanks! – 2017-01-14
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    @YourAverageEuler In finished proofs you never see all the failed attempts, so one might be tempted to think that in order to do well in mathematics you have to always be able to see the solution immediately. That is not the case at all. The best mathematicians in the world try and fail all the time, but it's the working solutions they share with the rest of the world. Of course, even their failed attempts are much better than what I could ever come up with, which is why I'm not a world-class mathematician. At least not yet :P – 2017-01-14
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Assume that $t^3$ does equal $ut+v$ (where $u=4$, $v=3$ for the first part, etc.). Then $t$ is a root of the cubic polynomial $$f(X)=X^3-uX-v \in\Bbb Z[X]$$ as well as of the quadratic polynomial $$g(X)=X^2-aX-b \in\Bbb Z[X].$$ But then $t$ is also a root of $$f(X)-Xg(X) =aX^2+(b-u)X-v$$ and also of $$h(X)=(f(X)-Xg(X))-ag(X)=(a^2+b-u)X+ab-v. $$ There are two possibilities:

  • If $a^2+b=u$ then we conclude $ab=v$. Two of the problem parts allow you to find $a,b$ with these properties. As the corresponding $g(X)$ has two real roots (the discriminant $a^2+4b$ is certainly positive), in these cases it is possible that $t^3=ut+v$.

  • If $a^2+b\ne u$, $h(X)$ has a unique root $t$ and it is rational. By the rational root theorem, $t$ must in fact be an integer and among the signed divisors of $v$. Conveniently, $v$ is prime in all problem parts, you you need only check for $t\in\{v,1,-1,-v\}$ whether or not $t^3=ut+v$. That is, check if $v^2=u+1$ or $u=0$ or $v=u-1$ or $v^2=u-1$.

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    Nice one @HagenvonEitzen This might be a general solution (like if the options were not given) is it? – 2017-01-14