In the equation $\sqrt{a}+\sqrt{b}+\sqrt{c}+\sqrt{d}+\sqrt{e}=\sqrt{f}$, each variable is a distinct positive integer. What is the least possible value for $f?$

Out of purely trial and error, I have come to the solution of a = 1, b= 4, c= 9, d = 16, e = 25 which sums to 15 thus $\sqrt{f} = 15$ and $f = 225$.

I suspect the final answer will turn out to be $f = 225 = 15^2$, but I have not yet managed to come up with a rigorous proof that no smaller value of $f$ is possible.

The main source of inspiration I've been trying work from so far is to consider a simpler variation of this problem where you only have two square roots on the left hand side, $$\sqrt{a}+\sqrt{b} = \sqrt{f}.$$ In this variation of the problem, we can square both sides to find that $$a+b+2\sqrt{ab} = f.$$ Therefore, $$2\sqrt{ab} = f-a-b,$$ consequently, $$4ab = (f-a-b)^2.$$ Since $a,b,f$ are integers and the left hand side ($4ab$) is an integer that is divisible by $2$, we see that the product $(f-a-b)^2 = (f-a-b)\cdot (f-a-b)$ is divisible by $2$. A product of integers is only divisible by a prime $p$ if and only if at least one of the factors is divisible by $p$. (This is a simple consequence of the Fundamental Theorem of Arithmetic, which is just the official name for the fact that prime factorizations of integers are unique.) Therefore, the integer factor $f-a-b$ itself must be divisible by $2$. In other words, we have deduced that $k = \frac{f-a-b}{2}$ must be an integer.

Thus we see that $$ab = k^2$$ is the square of an integer. Therefore, as a consequence of the Fundamental Theorem of Arithmetic (which is just the official name for the fact that prime factorizations of integers are unique), it must be the case that there exist integers $\alpha, \beta,\gamma$ such that $a = \alpha^2 \gamma$ and $b = \beta^2 \gamma,$ so \begin{align*}f &= a+b+2\sqrt{ab}\\ &= \alpha^2\gamma + \beta^2\gamma+2\sqrt{\alpha^2\beta^2\gamma^2}\\ &= (\alpha^2+\beta^2+2\alpha\beta)\gamma\\ &= (\alpha^2+\beta^2)\gamma\end{align*} Thus, in this simpler version of the problem where there is only a sum of two square roots, clearly the smallest possible value of $f$ comes from taking $\gamma = 1, \alpha = 1, \beta = 2$ which gives the solution $$\sqrt{1}+\sqrt{4} = \sqrt{9}.$$

I've been looking for ways to then bootstrap this argument (or something like it) up to something that can be applied to solving the original problem with a sum of five square roots. So far, I haven't found the right path forward.

I have also been looking for ways to exploit some routine square root inequalities. In particular, for any positive numbers $a_1, a_2,\dots,a_n$, it is the case that $$\sqrt{a_1+a_2+\dots+a_n} < \sqrt{a_1}+\sqrt{a_2}+\dots+\sqrt{a_n} \le \sqrt{n} \sqrt{a_1+a_2+\dots+a_n}.$$ This also has not yet led to progress.

I am consequently stuck here with no ideas how to continue.

Your help is appreciated!

Also, can you also help me on this($N$'s base-5 and base-6 representations, treated as base-10, yield sum $S$. For which $N$ are $S$'s rightmost two digits the same as $2N$'s?) problem too?




1 Answer 1


You are asking to determine the minimum $f$ in

$$\sqrt{a} + \sqrt{b} + \sqrt{c} + \sqrt{d} + \sqrt{e} = \sqrt{f} \tag{1}\label{eq1}$$

where each variable is a distinct positive integer. Change this slightly to

$$\sqrt{a} + \sqrt{b} + \sqrt{c} + \sqrt{d} + \sqrt{e} - \sqrt{f} = 0 \tag{2}\label{eq2}$$

Each integer has a maximum square-free factor, even if it's $1$, and a remaining perfect square factor, even if it's $1$. The perfect square factors can be taken out of the square root. Also, you can combine any terms which have the same square-free integer into one term with an integral multiplier. Thus, you get some $1 \le n \le 6$, with a result of

$$\sum_{i = 1}^n g_i \sqrt{h_i} = 0 \tag{3}\label{eq3}$$

for some integers $g_i$ and distinct square-free integers $h_i$.

In the MSE The square roots of different primes are linearly independent over the field of rationals, the answer by Qiaochu Yuan gives a link to his blog post Square roots have no unexpected linear relationships which says

IMO medalist Iurie Boreico has an article in an issue of the Harvard College Mathematics Review about his favorite problem:

Let $n_1, ... n_k$ be distinct squarefree integers. Show that if $a_1, ... a_k \in \mathbb{Z}$ are not all zero, then the sum $S = a_1 \sqrt{n_1} + ... + a_k \sqrt{n_k}$ is nonzero.

In other words, the problem is to show that there are no "unexpected" linear relationships between the fractions $\sqrt{1}, \sqrt{2}, \sqrt{3}, \sqrt{5}, \sqrt{6}, \ldots$.

If you can prove this problem, then all of the square-root values must be integral as there's only one subtraction, there can be just one term for the coefficient(s) to be $0$, so anything other than integral values would cause a larger value of $f$. Thus, the minimum value of the RHS is 15, resulting in $f = 225$, as you've already determined & conjectured. Unfortunately, the provided links don't work. Although Qiaochu gives a more sophisticated proof in that blog post, he also provides a working link in his MSE answer to Iurie's article. It is Harvard College Mathematics Review, Vol. 2, No. 1, Spring 2008, with the article starting at page 87.

  • 1
    $\begingroup$ @Max0815 You are welcome. As you can see, both proofs are both fairly long & complicated (at least they are to me), but I trust you can understand them and will find them useful, possibly even interesting. $\endgroup$ Feb 15, 2019 at 3:03
  • $\begingroup$ Yes. They are. I will have never thought of that way. I'm pretty sure I get all the parts but after I analyze it closely and try a similar problem I'll ask if I'm confused. $\endgroup$
    – Max0815
    Feb 15, 2019 at 3:05

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