Using fractions($f_n$) where integer $n$ is the number of fractions, prove that the sum of the squares $f_1$ to $f_n$ has no integer($I$) solutions when $n >= 1$ given each fraction has a distinct positive integer numerator and denominator and every fraction is a fully simplified non integer:

$$\left(f_1\right)^2 + \left(f_2\right)^2 + \left(f_3\right)^2 ... = I^2$$

I could not find a proof, but using distinct positive integers $a,b,c,d,e,f$ I managed to prove this for the case of $$\left(\frac{a}{b}\right)^2 + \left(\frac{c}{d}\right) ^2 = \left(\frac{e}{f}\right)^2 $$

by simplifying:

$$\frac{(ad)^2 + (bc)^2}{(bd)^2} = (\frac{e}{f})^2 $$

Since $(ad)^2 + (bc)^2$ must satisfy a Pythagorean triple, we know that each side length(excluding the hypotenuse) must be the product of two distinct integers, which can be proven through Euclid's formula :

$$ac = (2m)(n)$$ $$bd = (m -n)(m + n)$$


Although this still did not prove useful to answering my question, I later found a proof that the sum of the sqaures $f_1$ to $f_2$ has no integer solutions using distinct positive integers $a,b,c,d$ where $(\frac{a}{b})^2 + (\frac{c}{d} ) ^2 = I^2 $ using the following:

$$\frac {a^2}{b^2} = \frac {I^2d^2 - c^2}{d^2}$$

Both sides are in their simplest form since $$\gcd(a^2,b^2) = 1 = \gcd(c^2,d^2) = \gcd(I^2d^2 - c^2, d^2)$$

so $a^2 = e^2d^2-c^2, b^2 = d^2$.

Given $b,d > 0$, $b=d$. Therefore no solutions exists.

But this lead me to question what if there were more than $2$ integer fractions? This would mean that the denominator of some integer fraction could be the product of $2$ or more distinct integers, which suggests that this proof which uses $b = d$ does not nesscacarily need to be true for there to be an integer solution where $n > 2$.

So I made a conjecture that there are no integer solutions for the sum of any number of squared fractions where the numerator and denominator are distinct positive integers and the fractions are fully simplified non-integers.


This turned out to be false,

New questions:

Provide a proof

A method to finding integer solutions

(The one that finds the most affective method will be more likely to receive the bounty)

  • 1
    $\begingroup$ Could you elaborate on the conditions? The way they are stated is vague and according to how they are stated, this conjecture is obviously false for all $n\geq 1$, and even $0<n<1$. $\endgroup$ Commented Mar 8, 2021 at 19:38
  • $\begingroup$ @Alexander Conrad, Could you please expand on what exactly seems vague?I also have already provided a proof that there are no integer solutions where $n = 2$ and am quite confused by what you mean when you say $0 < n < 1$ as $n$ is always an integer. $\endgroup$ Commented Mar 8, 2021 at 19:47
  • 1
    $\begingroup$ If you write $f_i = a_i / b_i$ with $\gcd(a_i, b_i) = 1$, what are the conditions on the $a_i$ and $b_i$? All $f_i$ must not be integers? And all the $a_i$'s have to be different? And all the $b_i$'s as well? Please clarify. $\endgroup$ Commented Mar 8, 2021 at 20:09
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    $\begingroup$ @Bart Michels, The conditions for $a_i $and $b_i$ are exactly as you say as well as for $ f_i$ as long as $a_i$ and $b_i $ are integers. $\endgroup$ Commented Mar 8, 2021 at 20:12
  • $\begingroup$ I assume we won't count $n=1$, $f_1 = (-1)/1$. $\endgroup$
    – aschepler
    Commented Mar 8, 2021 at 20:17

2 Answers 2


I did a brute force search in sage to find examples of the form $$\left(\frac{x}{qr}\right)^2 + \left(\frac{y}{pr}\right)^2 + \left(\frac{z}{pq}\right)^2= n^2$$ with $p, q, r$ primes that are $1$ mod $4$. The condition of this being an integer more or less fixes $x$ and $y$ mod $r^2$ (because $x^2/q^2+y^2/p^2$ must be divisible by $r^2$, and there are only 8 interesting solutions to that). Similarly it fixes $y$ and $z$ mod $p^2$, and $x$ and $z$ mod $q^2$. This gives a system of congruences in $x, y, z$ mod $p^2q^2r^2$. Then brute force search through $p, q, r$ and solutions to the system of congruences by adding multiples of $p^2q^2r^2$ to $x, y, z$, and hope that it gives perfect squares. (I did not but could have separately added multiples of $q^2r^2$ to $x$, of $p^2r^2$ to $y$ and of $p^2q^2$ to $z$. I just translated all of them by $m \cdot p^2q^2r^2$ for simplicity.)

There are plenty of solutions.

The smallest $n$ I saw is $81$:

$(286448/4141)^2 + (1636/205)^2 + (20897/505)^2 = 81^2$

Other solutions (I didn't filter for duplicates)

(221622/481)^2 + (10777/185)^2 + (3186/65)^2 = 467^2

(15246/533)^2 + (28777/205)^2 + (3102/65)^2 = 151^2

(75528/533)^2 + (13248/205)^2 + (2898/65)^2 = 162^2

(11943/793)^2 + (25992/305)^2 + (2848/65)^2 = 97^2

(1188973/1469)^2 + (217148/565)^2 + (2764/65)^2 = 897^2

(332352/1937)^2 + (88104/745)^2 + (1039/65)^2 = 209^2

(1715638/2509)^2 + (819003/965)^2 + (3186/65)^2 = 1091^2

(5821714/2561)^2 + (39784/985)^2 + (2848/65)^2 = 2274^2

(1785867/2977)^2 + (1258284/1145)^2 + (3236/65)^2 = 1253^2

(231632/493)^2 + (941/145)^2 + (6976/85)^2 = 477^2

(151859/1513)^2 + (71094/445)^2 + (2526/85)^2 = 191^2

(498166/1717)^2 + (224522/505)^2 + (907/85)^2 = 531^2

(1723985/2669)^2 + (345746/785)^2 + (3682/85)^2 = 783^2

(231632/493)^2 + (6976/85)^2 + (941/145)^2 = 477^2

(650784/3277)^2 + (204864/565)^2 + (20084/145)^2 = 436^2

(221622/481)^2 + (3186/65)^2 + (10777/185)^2 = 467^2

(80758/4181)^2 + (216998/565)^2 + (8039/185)^2 = 387^2

(3610252/8473)^2 + (1258884/1145)^2 + (12261/185)^2 = 1181^2

(15246/533)^2 + (3102/65)^2 + (28777/205)^2 = 151^2

(75528/533)^2 + (2898/65)^2 + (13248/205)^2 = 162^2

And one where none of the primes is $5$:

(155476/1241)^2 + (164575/949)^2 + (33040/221)^2 = 261^2

  • $\begingroup$ Thanks, your examples helped spot my error: the denominators as written in one expression were distinct, but they aren't necessarily the correct lowest-form denominators. The actual least denominators could have duplicates, making the reduction to smaller solution argument invalid. $\endgroup$
    – aschepler
    Commented Mar 8, 2021 at 21:54
  • $\begingroup$ @aschepler Regarding your deleted answer: Nice argument, but it's not true that all $b_i$ are distinct integers. In fact, after reducing $\frac{a_i \operatorname{lcm}(\beta^c)}{b_i}$ to lowest terms, we might end up with equal denominators.This is exactly what is used in Bart Michael's answer. Nevertheless, I'd leave your answer since it shows what the structure of the denominators must be in order to have a solution. $\endgroup$
    – Kolja
    Commented Mar 8, 2021 at 21:57
  • $\begingroup$ @Kolja It's not obvious to me how to explain or apply that, though I do see a relation to how Bart's choice of $qr, pr, pq$ helps make a solution possible (and breaks my argument). I'll make it visible again for now for further discussion. $\endgroup$
    – aschepler
    Commented Mar 8, 2021 at 22:03

This proof is INVALID, but pieces of it might give insights to the forms of solutions...

In fact as long as just the denominators are distinct, the sum of squares is never an integer, so it can't be the square of an integer.

Let's call the fractions $\frac{a_i}{b_i}$, where $a_i, b_i \in \mathbb{N}, b_i \geq 2$, and we require that for each index $i$, $\gcd(a_i,b_i)=1$ and for all index pairs $i \neq j$ that $b_i \neq b_j$. We look for solutions where:

$$S = \sum_{i=1}^n \frac{a_i^2}{b_i^2} \in \mathbb{Z}$$

Suppose by way of contradiction that the set of such solutions is not empty. Let $N$ be the smallest count with at least one solution using exactly $n=N$ fractions in this set. Then let $B$ be the smallest positive integer which appears as a denominator $b_i$ in any solution of size $N$.

Consider a "minimal" solution which has $n=N$ and has $b_i=B$ for some $i$. Let $p$ be a prime factor of $B$ (which is at least $2$). Let $\beta$ be the set of denominators $b_i$ which are a multiple of $p$. This is not the empty set since $p \mid B \Rightarrow B \in \beta$. If $\beta$ is the entire set $\{b_1, \ldots b_N\}$, then we have

$$ \sum_{i=1}^N \frac{a_i^2}{(b_i/p)^2} = p^2 \sum_{i=1}^N \frac{a_i^2}{b_i^2} = p^2 S \in \mathbb{Z} $$

is a solution with the same $N$, and one of the denominators is $B/p$, but this contradicts the choice of $B$ as the smallest denominator in any solution of size $N$.

If $\beta$ is not the entire set $\{b_1, \ldots b_N\}$, then the set difference $\beta^c = \{b_1, \ldots b_N\} \setminus \beta$ is not empty. The least common multiples of these two subsets of denominators satisfy $p \mid \mathop{\rm lcm}(\beta)$ and $p \nmid \mathop{\rm lcm}(\beta^c)$. Splitting the sum $S$ over these two subsets:

$$ S = \sum_{b_i \in \beta} \frac{a_i^2}{b_i^2} + \sum_{b_i \in \beta^c} \frac{a_i^2}{b_i^2} \\ S (\mathop{\rm lcm} \beta^c)^2 = \sum_{b_i \in \beta} \frac{a_i^2 (\mathop{\rm lcm} \beta^c)^2}{b_i^2} + \sum_{b_i \in \beta^c} a_i^2 \left(\frac{\mathop{\rm lcm} \beta^c}{b_i}\right)^2 \\ \sum_{b_i \in \beta} \left(\frac{a_i \mathop{\rm lcm}(\beta^c)}{b_i}\right)^2 = S (\mathop{\rm lcm} \beta^c)^2 - \sum_{b_i \in \beta^c} a_i^2 \left(\frac{\mathop{\rm lcm} \beta^c}{b_i}\right)^2 \in \mathbb{Z} $$

For each addend term in the left side, $p \mid b_i$ by definition of $\beta$. Since $\gcd(a_i, b_i) = 1$, $p \nmid a_i$, and we already noted $p \nmid \mathop{\rm lcm}(\beta^c)$. So the fraction is not an integer. Every $b_i$ in these fractions is a distinct integer, since they're a subsequence of the entire sequence of $N$ different $b_i$. Therefore the left side is a solution to the problem with $n = |\beta| < N$, contradicting the choice of $N$ as minimal.

EDIT: Here's the error. The $b_i$ values are distinct, but they are not necessarily the denominators of the new fractions when reduced to lowest form. The actual reduced denominators could duplicate some values, meaning the formula is not another solution to the problem as I modified it.

All possibilities lead to contradiction, so there are no solutions at all.

  • $\begingroup$ I haven't checked where your argument is wrong, but there are solutions. (See my answer.) $\endgroup$ Commented Mar 8, 2021 at 21:42

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