# Proving that if $\sum_{i =1}^ t {a_i} = \sum_{i=1}^ t {b_i}$, then $\sum_{i =1}^ t {a_i^2} \neq \sum_{i=1}^ t {b_i^2}$.

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 $$t > 2$$ ?

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

• I think you can do it with three instead of 4 May 17, 2021 at 7:49

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.