Showing that $a^2+b^2+c^2+d^2+e^2+65=abcde$ has integer solutions greater than $2018$?

This question comes from a Chinese high school olympiad training program. It seems remarkably more difficult (and indeed, interesting!) than all other problems arising in the same program, especially since an elementary (high-school level) solution is probably available.

Show that there exists integers $$a,b,c,d,e>2018$$ so that the equation $$a^2+b^2+c^2+d^2+e^2+65=abcde$$ is satisfied.

For what it's worth, here's what I have tried. Rewriting the equation as a quadratic polynomial in $$a$$, $$a^2-(bcde)a+b^2+c^2+d^2+e^2+65=0.$$ For there to be integer solutions, the discriminant must be a perfect square. Hence $$b^2c^2d^2e^2-4b^2-4c^2-4d^2-4e^2-4\cdot5\cdot13=n^2.$$ I don't however see how I can solve this equation, especially due to the large number of unknowns. Any ideas?

Edit: Ivan Neretin presents an excellent answer by Vieta Jumping, which I'm sure will yield results. However, the training program I mentioned has not discussed such advanced tactics as Vieta Jumping yet, and only covered $$\gcd$$, $$\operatorname{lcm}$$, factorisation of polynomials, discriminants, modular arithmetic, divisibility and quadratic residues. Hence despite Ivan's excellent solution, I would still be extremely appreciative of a more elementary solution.

• Why is $abcde$ not orders of magnitude greater than the left hand side of the equation? If $a, b, c, d, e > 2018$ then the left hand is about $10^7$ whereas the right hand side should be about $10^{16}$, no? – stuart stevenson Nov 26 '18 at 23:00
• @stuartstevenson: Only if all of $a,b,c,d,e$ are similarly-sized. – Micah Nov 26 '18 at 23:05
• @Micah So you're saying that $a$, for instance, is orders of magnitude greater than $e$ assuming $a \geq b \geq c \ldots$? – stuart stevenson Nov 26 '18 at 23:10
• @stuartstevenson Yes, we can have that. If we set $b=c=d=e$ for example we have a quadratic in $a$ which clearly yields a solution. The same is true for any other fixed values of $b,c,d,e$; we will always have a solution for $a$, even if orders of magnitude larger. The question does not care about the relative sizes of $a,b,c,d,e$ at all. – YiFan Nov 26 '18 at 23:13
• Following Ivan Neretin's answer, I get following monstrous solution: \begin{align} a &= 7138\\ b &= 16988437\\ c &= 72151760667066\\ d &= 1041175313471572184867943319\\ e &= 9109630532627114315851511163018235051842553960810405 \end{align} – achille hui Nov 26 '18 at 23:18

You are supposed to bruteforce or guess $$(a,b,c,d,e)=(1,2,3,4,5)$$ or any other small solution, and then go up by Vieta jumping. That is, once you have a solution, you rewrite it as a quadratic polynomial in $$a$$ (just like you did), and since one root is integer, so is the other. Then you do the same to $$b,\;c...$$ and repeat until the roots are big enough.

• great answer, I tried it out and got a=9109630532627114315851511163018235051842553960810405,b=1041175313471572184867943319,c=7138,d=16988437,e=72151760667066; – miracle173 Nov 26 '18 at 23:18
• Unfortunately the training program has not covered Vieta Jumping yet; it is at a much more elementary level (see the question edit). I'll accept this great answer if none else comes by after a day, but meanwhile, are you aware of a more elementary solution? – YiFan Nov 26 '18 at 23:28
• @YiFan Maybe you should not use Vieta Jumping but only the fact that $x^2+px+q=(x-x_1)(x-x_2)$ if $x_1$ and $x_2$ are the solutions of $x^2+px+q=0$ – miracle173 Nov 26 '18 at 23:46
• @miracle173 I'm aware of that fact, but how does that help? – YiFan Nov 27 '18 at 0:00
• @YiFan If $(a,b,c,d,e)$ is a solution with $0 < a < b < c < d < e$, then $(b,c,d,e,bcde-a)$ is another solution with $0 < b < c < d < e < bcde - a$. Start from the seed solution $(1,2,3,4,5)$, this allow us to construct a sequence of solutions whose smallest member is increasing. I think this is something the kids can understand. – achille hui Nov 27 '18 at 0:15

Here is the output of the Maxima commands I used to calculate a solution according to Ivan Neretin's answer.

(%i2) ev(x^2-b*c*d*e*x+b^2+c^2+d^2+e^2+65,x = 1,b = 2,c = 3,d = 4,e = 5)

(%o2)                                  0

(%i3) ev(solve(x^2-b*c*d*e*x+b^2+c^2+d^2+e^2+65,x),b = 2,c = 3,d = 4,e = 5)

(%o3)                          [x = 1, x = 119]

(%i4) ev(solve(x^2-b*c*d*e*x+b^2+c^2+d^2+e^2+65,x),b = 119,c = 3,d = 4,e = 5)

(%o4)                          [x = 7138, x = 2]

(%i5) ev(solve(x^2-b*c*d*e*x+b^2+c^2+d^2+e^2+65,x),b = 119,c = 7138,d = 4,e

= 5)

(%o5)                        [x = 3, x = 16988437]

(%i6) ev(solve(x^2-b*c*d*e*x+b^2+c^2+d^2+e^2+65,x),b = 119,c = 7138,

d = 16988437,e = 5)

(%o6)                     [x = 72151760667066, x = 4]

(%i7) ev(solve(x^2-b*c*d*e*x+b^2+c^2+d^2+e^2+65,x),b = 119,c = 7138,

d = 16988437,e = 72151760667066)

(%o7)              [x = 1041175313471572184867943319, x = 5]

(%i8) ev(solve(x^2-b*c*d*e*x+b^2+c^2+d^2+e^2+65,x),

b = 1041175313471572184867943319,c = 7138,d = 16988437,

e = 72151760667066)

(%o8) [x = 9109630532627114315851511163018235051842553960810405, x = 119]

(%i9) ev(x^2-b*c*d*e*x+b^2+c^2+d^2+e^2+65,

x = 9109630532627114315851511163018235051842553960810405,

b = 1041175313471572184867943319,c = 7138,d = 16988437,

e = 72151760667066)

(%o9)                                  0


This expands upon @Ivan Neretin 's answer

This is what is meant by Vieta jumping, when it comes to this problem--we can use Vieta jumping to prove the following claim:

Proposition 1: For any $$A \in \mathbb{Z}^+$$ there is a solution $$(a,b,c,d,e)$$; $$a,b,c,d,e \in \mathbb{Z}^+$$ to the equation $$a^2+b^2+c^2+d^2+e^2+65 = abcde$$ such that $$A \leq \min\{a,b,c,d,e\}$$.

Proof: We first claim the following:

Claim 1: Let us suppose that there is a solution $$a^2+b^2+c^2+d^2+e^2+65=120$$; $$a,b,c,d,e \in \mathbb{Z}^+$$, and let us assume WLOG that $$a= \min\{a,b,c,d,e\}$$. Then there is a solution $$(a',b',c',d',e')$$; $$a',b',c',d',e' \in \mathbb{Z}^+$$ s.t. $$a'>a$$ and $$b'=b,c'=c,d'=d,e'=e$$.

[Proof of Claim 1: Indeed, iff there exists an integer $$a'> a$$ is that satisfies the following:

$$a'^2 + (b^2+c^2+d^2+e^2)+65 > a'(bcde)$$.

Then Claim 1 will follow. However, the above is a quadratic equation in $$a'$$ of the form $$a'^2 - xa' + z$$; both $$x$$ and $$z$$ positive integers, that has at least one integer solution, namely $$a$$, and so the other solution is an integer; as $$z$$ is at least $$b^2+c^2+d^2+e^2+65$$ $$> 4a^2$$ and $$aa'=z$$ with $$a$$ positive it follows that $$a'$$ must be strictly greater than $$4a$$. So indeed Claim 1 does follow. $$\surd$$ ]

We note that Proposition 1 follows immediately from Claim 1, and the existence of at least one solution $$(a_0,b_0,c_0,d_0,e_0); a_0,b_0,c_0,d_0,e_0 \in \mathbb{Z}^+$$ such that the equation $$a_0^2+b_0^2+c_0^2+d_0^2+e_0^2+65 = a_0b_0c_0d_0e_0$$ holds; namely $$(a_0,b_0,c_0,d_0,e_0) =(1,2,3,4,5)$$. $$\surd$$