mruczek wrote: Numbers $a, b, c$ are such that $3a + 4b = 3c$ and $4a - 3b = 4c$. Show that $a^2 + b^2 = c^2$. The two equations imply $b=0$ and $a=c$. Hence the conclusion

Faster solution. Both equations rise to the second power and add sides. You'll get $25a^2+25b^2=25c^2$.

This problem was proposed by Burii.

This is the first olympiad problem I've seen that I can solve

From the first equation we have a+4b/3=c .And we have 3a+4b+4a-3b=7c,7a+b=7c .Then from 7a+b=7c we have a+ b/7=c .Then c-a=4b/3= b/(7 ) , 4b/3= b/7 28b=3b b=0 Then we have 3a+0=3c.,a=c. a^2+ b^2= c^2+ 0= c^2

Solve the two equations and we can get a=c, b=0. So a²+b²=c²

Numbers $a, b, c$ are such that $3a + 4b = 3c$ and $4a - 3b = 4c$. Show that $a^2 + b^2 = c^2$. Note that a few solutions are$$(1,0,1), (2,0,2), (3,0,3), ...$$so we then make the following claim: Claim: The only such solutions are in the form $(a,b,c)=(a_1,0,a_1),$ and in other words, $a=c, b=0.$ Proof: We manipulate the equations to get$$7a+b=7c$$$$a+7b=c$$from which we see that $a=c, b=0.$ $~\blacksquare$ Thus, since $a=c$ and $b=0,$$$a^2+b^2=c^2,$$as desired. $~\blacksquare$

Solve the two equations and we get $a=c, b=0$ and we are done.

$4b=3(c-a)$ and $-3b=4(c-a)$. This implies that $c-a=0$, so $b=0$ and $a=c$. Therefore, $a^2+b^2=c^2$.

Notice that by little work, "plug and chug" you can come up with the following equality $$ 7a = 24b $$ Now notice that $$ a^{2} -14ab + 49b^{2} = c^{2} $$$$ a^{2} - 2b(7a -24b) + b^{2} = c^{2} $$$$ \implies a^2 + b^2 = c^2 $$

$\color{blue} \boxed{\textbf{SOLUTION}}$ $(3a+4b)^2=(3c)^2 \implies 9a^{2}+24ab+16b^{2}=9c^2$ And, $(4a-3b)^2=(4c)^2 \implies 16a^{2}-24ab+9b^{2}=16c^{2}$ Adding these gives, $25a^{2}+25b^{2}=25c^{2} \implies a^{2}+b^{2}=c^{2} \blacksquare$

$$3a+4b=3c \quad (1)$$$$4a-3b=4c \quad (2)$$ Firstly (to eliminate $b$) we multiply (1) by $3$ and (2) by $4$,then we have: $$ \Rightarrow 9a + 12b = 9c \quad (1.1), $$$$ \Rightarrow 16a - 12b = 16c \quad (1.2.) $$$$(1.1)+(1.2) \Rightarrow 25a=25c \Rightarrow a=c \Rightarrow a^2=c^2$$Now, (to eliminate $a$) we multiply (1) by $4$ and (2) by $-3$,then we have: $$ \Rightarrow 12a+16b=12c\quad (2.1)$$$$ \Rightarrow -12a+9b=-12c\quad (2.2)$$$$(2.1)+(2.2) \Rightarrow 25b=0 \Rightarrow b=0 \Rightarrow b^2=0$$Finally we have that: $$a^2+b^2=c^2+0=c^2$$Q.E.D

Numbers $a, b, c$ are such that $ a + 2b = c$ and $ 2a - b =2c$. Show that $$a^2 + b^2 = c^2$$Numbers $a, b, c$ are such that $3a + 2b = 3c$ and $ 2a - 3b =2c$. Show that $$a^2 + b^2 = c^2$$