1
$\begingroup$

Suppose $t > 0$ is an integer, $ a_i >0 $ and $ b_i >0$ are all integer, where $i \in \{1, 2, \ldots, t \}$. Suppose all $a_i$ and $b_j$, $i, j \in \{1, 2, \ldots, t \}$, are pairwise distinct. Whether the following claim is true for any positive integer $t $ ?

If $\sum_{i =1}^ t {a_i} = \sum_{i=1}^ t {b_i}$, then $\sum_{i =1}^ t {a_i^2} \neq \sum_{i=1}^ t {b_i^2}$.

If $t=2$, the claim can be describe as follows. If $a_1 + a_2 = b_1 + b_2 $, then $ a_1^2 + a_2^2 \neq b_1^2 + b_2^2 $.

In this case without loss of generality, we may assume that $a_1 < a_2$, $b_1 < b_2 $ and $a_1 > b_1$. This implies $a_1 + b_1 < a_2 + b_2$. In addition, as $a_1 + a_2 = b_1 + b_2$, $a_1 - b_1 = b_2 - a_2$. Therefore, $a_1^2 - b_1^2 = (a_1 + b_1) (a_1 - b_1 ) = (a_1 + b_1) (b_2 - a_2) < (a_2 + b_2) (b_2 - a_2) = b_2^2 - a_2^2$. Accordingly, $ a_1^2 + a_2^2 < b_1^2 + b_2^2 $. Hence, the claim is true in the case of $t =2$. However, the claim must be correct when $t > 2$ ?

$\endgroup$
0

2 Answers 2

1
$\begingroup$

An oldie, compare $1,4,6,7$ with $2,3,5,8$

$\endgroup$
2
  • $\begingroup$ Thanks for your answer ! $\endgroup$
    – Mike. G
    May 17, 2021 at 7:42
  • $\begingroup$ I think you can do it with three instead of 4 $\endgroup$
    – Empy2
    May 17, 2021 at 7:49
1
$\begingroup$

For the case $t=2$, we can prove a more general result . . .

Claim:

If $K$ is a field with $\text{char}(K)\ne 2$, and if $a_1,a_2,b_1,b_2\in K$ are such that \begin{align*} a_1+a_2&=\,b_1+b_2 \qquad\qquad\qquad\qquad\qquad\qquad\;\;\;\;\; \\[4pt] a_1^2+a_2^2&=\,b_1^2+b_2^2\\[4pt] \end{align*} then $\{a_1,a_2\}=\{b_1,b_2\}$.

Proof:

\begin{align*} & \begin{cases} a_1+a_2&\!\!\!\!=\,b_1+b_2\\[4pt] a_1^2+a_2^2&\!\!\!\!=\,b_1^2+b_2^2\\[4pt] \end{cases} \\[4pt] \implies\;& (a_1+a_2)^2-(a_1^2+a_2^2)=(b_1+b_2)^2-(b_1^2+b_2^2) \\[4pt] \implies\;& 2a_1a_2=2b_1b_2 \\[4pt] \implies\;& a_1a_2=b_1b_2 && \!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\! \bigl(\text{since char$(K)\ne 2$}\bigr) \\[4pt] \implies\;& (x-a_1)(x-a_2)=(x-b_1)(x-b_2) && \!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\! \bigl(\text{as elements of $K[x]$}\bigr) \\[4pt] \implies\;& \{a_1,a_2\}=\{b_1,b_2\} && \!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\! \bigl(\text{since $K[x]$ is a UFD}\bigr) \\[4pt] \end{align*} as was to be shown.

However for the case $t=3$, there are counterexamples.

Here are two such counterexamples: \begin{align*} 1+5+6&=2+3+7\\[4pt] 1^2+5^2+6^2&=2^2+3^2+7^2\\[14pt] 1+6+8&=2+4+9\\[4pt] 1^2+6^2+8^2&=2^2+4^2+9^2\\[4pt] \end{align*} Moreover, based on limited testing, the following conjecture appears likely to hold:

Conjecture:

For any positive integer $k > 1$, there exist infinitely many $4$-tuples $(a_2,a_3,b_2,b_3)$ of positive integers with $1 < a_2 < a_3$ and $k < b_2 < b_3$ such that $$ \{a_2,a_3\}\cap\{k,b_2,b_3\}={\large{\varnothing}} $$ and such that \begin{align*} 1+a_2+a_3&=k+b_2+b_3\\[4pt] 1^2+a_2^2+a_3^2&=k^2+b_2^2+b_3^2\\[4pt] \end{align*}

Update:$\;$The conjecture is true.

Proof:

Let $k\;$be a positive integer with $k > 1$, let $u\;$be a positive integer with $u > 3k-2$, and let $a_2,a_3,b_2,b_3$ be given by \begin{align*} a_2&=u\\[2pt] a_3&=2u+2-3k\\[9pt] b_2&=u+2-2k\\[2pt] b_3&=2u+1-2k\\[4pt] \end{align*} From $$ \left\lbrace \begin{align*} &a_2-1=u-1 > (3k-2)-1=3(k-1) > 0\\[4pt] &a_3-a_2=(2u+2-3k)-u=u+2-3k > (3k-2)+2-3k=0 \qquad\qquad\;\, \\[4pt] \end{align*} \right. $$ we get$\;1 < a_2 < a_3$.

From $$ \left\lbrace \begin{align*} &b_2-k=(u+2-2k)-k=u+2-3k > (3k-2)+2-3k=0\\[4pt] &b_3-b_2=(2u+1-2k)-(u+2-2k)=u-1 > (3k-2)-1=3(k-1) > 0\\[4pt] \end{align*} \right. $$ we get$\;k < b_2 < b_3$.

From $$ \left\lbrace \begin{align*} &a_2-k=u-k > (3k-2)-k=2(k-1) > 0\\[4pt] &a_2-b_2=u-(u+2-2k)=2(k-1) > 0\\[4pt] &a_2-b_3=u-(2u+1-2k)=2k-1-u < 2k-1-(3k-2)=-(k-1) < 0\\[4pt] \end{align*} \right. $$ and $$ \left\lbrace \begin{align*} &a_3-k=(2u+2-3k)-k=2u+2-4k > 2(3k-2)+2-4k=2(k-1) > 0\\[4pt] &a_3-b_2=(2u+2-3k)-(u+2-2k)=u-k > (3k-2)-k=2(k-1) > 0\\[4pt] &a_3-b_3=(2u+2-3k)-(2u+1-2k)=-(k-1) < 0\\[4pt] \end{align*} \right. $$ we get$\;\{a_2,a_3\}\cap\{k,b_2,b_3\}={\large{\varnothing}}$.

Finally, from $$ \left\lbrace \begin{align*} &(1+a_2+a_3)-(k+b_2+b_3)\\[4pt] &=\Bigl(1+u+(2u+2-3k)\Bigr)-\Bigl(k+(u+2-2k)+(2u+1-2k)\Bigr) \qquad\qquad\;\;\;\;\;\; \\[4pt] &=0\;\;\;\bigl(\text{after expanding and collecting like terms}\bigr)\\[4pt] \end{align*} \right. $$ and $$ \left\lbrace \begin{align*} &(1^2+a_2^2+a_3^2)-(k^2+b_2^2+b_3^2)\\[4pt] &=\Bigl(1^2+u^2+(2u+2-3k)^2\Bigr)-\Bigl(k^2+(u+2-2k)^2+(2u+1-2k)^2\Bigr) \qquad\;\;\;\; \\[4pt] &=0\;\;\;\bigl(\text{after expanding and collecting like terms}\bigr)\\[4pt] \end{align*} \right. $$ we get $$ \left\lbrace \begin{align*} 1+a_2+a_3&=k+b_2+b_3\\[4pt] 1^2+a_2^2+a_3^2&=k^2+b_2^2+b_3^2 \qquad\qquad\qquad\qquad\qquad\qquad \\[4pt] \end{align*} \right. $$ hence, since for a given value of $k$, the parameter $u\;$can be chosen arbitrarily large, the specified parametric form yields infinitely many $4$-tuples $(a_2,a_3,b_2,b_3)$ satisfying the required conditions.

This completes the proof of the conjecture.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .