Can square roots be simplified without a calculator?

[littlemathquark]

Homework Statement

$\sqrt{\dfrac{3^8+5^8+34^4}2} =?$

Relevant Equations

None

I can't find a short solution without using calculator.

 

[Halc]

Consider the prime factors of all the components
Oh, that only works if the 3 upper terms are multiplied.
Adding them makes only for a big unwieldy number. Don't see how to simplify it at all.

 

[Ibix]

Do 3, 5, and 34 (or their prime factors) have any properties in common that might help? Especially helpful if there are terms that might cancel when expanded.

 

[littlemathquark]

Do 3, 5, and 34 (or their prime factors) have any properties in common that might help? Especially helpful if there are terms that might cancel when expanded.

$3^8+5^8+34^4=9^4+25^4+34^4=9^4+25^4+(9+25)^4$
That's all and I'm stuck.

 

[Ibix]

Interesting! Not what I had in mind at all, but it also works.

What do you get if you expand that? I found it helpful to write $x=3$ and $y=5$ so I didn't lose track of my special numbers among all the other ones.

If the expansion is going to help you in this particular problem, what has to be true about it?

 

[fresh_42]

$3^8+5^8+34^4=9^4+25^4+34^4=9^4+25^4+(9+25)^4$
That's all and I'm stuck.

This is a good starting point! Just go ahead by this method.
\begin{align*}
3^8+5^8+34^4&=9^4+25^4+(9+25)^4\\
&=9^4+(16+9)^4+(9+16+9)^4\\
&=9^4+(2\cdot 9+7)^4+(3\cdot 9+7)^4\\
&\ldots
\end{align*}
I put $9^4$ out of it and called $7/9=c$ and then $2+c=25/9=d$ and so on. The highest product I finally had to calculate was $25\cdot 34=25(25+9)=625+225.$

 

[littlemathquark]

This is a good starting point! Just go ahead by this method.
\begin{align*}
3^8+5^8+34^4&=9^4+25^4+(9+25)^4\\
&=9^4+(16+9)^4+(9+16+9)^4\\
&=9^4+(2\cdot 9+7)^4+(3\cdot 9+7)^4\\
&\ldots
\end{align*}
I put $9^4$ out of it and called $7/9=c$ and then $2+c=25/9=d$ and so on. The highest product I finally had to calculate was $25\cdot 34=25(25+9)=625+225.$

Sorry I don't understand your solution. Can you give more detail please?

 

[fresh_42]

Sorry I don't understand your solution. Can you give more detail please?

\begin{align*}
3^8+5^8+34^4&=9^4+25^4+(9+25)^4\\
&=9^4+(16+9)^4+(9+16+9)^4\\
&=9^4+(2\cdot 9+7)^4+(3\cdot 9+7)^4\\
&=9^4\cdot (1+(2+c)^4+(1+2+c)^4)\ ,\ c=7/9\\
&=9^4\cdot (1+d^4+(1+d)^4)\, , \,d=2+c=25/9\\
&=9^4\cdot \left(1+d^4+d^4+4d^3+6d^2+4d+1\right)\\
&=9^4\cdot \left(2+2d^4+4d^3+6d^2+4d\right)\\
\dfrac{3^8+5^8+34^4}{2}&=9^4\cdot \left(1+d^4+2d^3+3d^2+2d\right)\\
&\ldots
\end{align*}
The only critical part now is figuring out that $1+d^4+2d^3+3d^2+2d=1+d^4+d^2+2d^3+2d^2+2d$ is a square. I admit, I used WA for it, but more out of laziness.

 

[PeroK]

Here's the key:
$$m^4 + (m+ k)^4 + (2m+k)^4 = 2(3m^2 + 3km + k^2)^2$$E.g. with $m = 9,k = 16$, we have:
$$9^4 + 25^4 + 34^4 = 2(243 + 432 + 256)^2 = 2(931)^2$$

 

[Vanadium 50]

That seems kind of out of the blue.

My starting point would be to replcae 3 with 4-1, 5 with 4+1 and use that to get rid of the 2. Then see what I had left.

 

[Ibix]

That seems kind of out of the blue.

My starting point would be to replcae 3 with 4-1, 5 with 4+1 and use that to get rid of the 2. Then see what I had left.

That's what I did, additionally noting that $34=2(4^2+1)$. That gave me a polynomial expression in powers of 4 that can be factorised (and there's an obvious factorisation that we want). OP's approach with $3$, $5$, and $34=3^2+5^2$ can be made to work similarly.

 

[PeroK]

For any integers $m, n$ the sum $m^4+n^4 + (m +n)^4$ is twice a perfect square. There's nothing special about 9 and 25.

The equation is easier to derive with $k = n -m$.

 

[PeroK]

This is why algebra was invented!

 

[littlemathquark]

Another solution

$$3^8+5^8+34^4=9^4+25^4+34^4=9^4+25^4+(9+25)^4$$

$$\frac12\left(a^4+b^4+(a+b)^4\right)=\frac12(a^4+b^4+a^4+4a^3b+6a^2b^2+4ab^3+b^4)$$

$$=(a^2)^2+(b^2)^2+(ab)^2+2(a^2b^2+a^3b+ab^3)=(a^2+b^2+ab)^2$$

$$(\forall a,b\in\mathbb{R}$$ $$a^2+b^2+ab\geq0 )$$,

$$\sqrt{\dfrac{a^4+b^4+(a+b)^4}2}=a^2+b^2+ab$$

$$ \sqrt{\dfrac{3^8+5^8+34^4}2}=9^2+25^2+9\cdot25=931$$