Non geos are cool part 2 and Mailchimp!

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And I am back with 5 more cute Non-geos!  So here are some 5 Non-geos that were cool! 😎 I think I should say them as NTs.

Problem: Determine all integers solutions of the equation $x + x^3 = 5y^2$.

Proof: Answer: $(x,y)=(0,0)$

We use infinite descent.

 

Let $(a,b)$ be the least solution, hence $\gcd (a,b)=1.$

So $a+a^3=a(a^2+1)=5b^2\implies a\equiv 0,2,3\mod 5.$

Taking $\mod 4$ gives us $$a(a^2+1)\equiv b^2\mod 4\implies a(a^2+1)\equiv 0,1\mod 4.$$

Clearly $$2|a(a^2+1)\implies a(a^2+1)\equiv 0\mod 4\implies 2|a.$$

So we get $$2|a+a^3=b\implies 2|\gcd(a,b)\implies \gcd(a,b)\ne 1.$$ 

Contradiction. And we are done by infinite descent.

So the only sol is $(0,0).$

Problem: Let $p> 3$ be a prime number and let $q = \frac{4^p-1}{3}$. Show that $q$ is a composite integer as well is a divisor of $2^{q-1}- 1$.

Proof: The first part was easy. We know that $4^p-1=(2^p-1)(2^p+1).$ We know that $3|2^p+1$ and $\frac{2^p+1}{3}>1.$ Hence $$q = \frac{4^p-1}{3}=(2^p-1)\cdot \frac{2^p+1}{3}.$$ So $q$ is composite.

Now for the second part. Well, we have $q-1=\frac{4^p-4}{3}=4\cdot \frac{4^{p-1}-1}{3}.$ By Fermat's little theorem, we have $p|4^{p-1}-1\implies p|q-1.$ So let $q-1=p\cdot l.$ Clearly $q-1>p\implies l>1.$

So, we have $$2^{q-1}-1=2^{p\cdot l}-1=(2^p)^l-1^l\implies \boxed{2^p-1|2^{q-1}-1}.$$

Now clearly since $q-1$ is even, $p$ is odd, hence $l$ is even $l=2m.$

Now, we want $$ \frac{2^p+1}{3}|2^{q-1}-1=4^{pm}-1=(2^{pm}-1)\cdot (2^{pm}+1).$$

Clearly $$\boxed{2^p+1|2^{q-1}-1}. $$

Since $\gcd(2^p-1,2^p+1)=1.$

We get that $$ q|2^p-1\cdot 2^p+1| 2^{q-1}-1.$$ We are done.

Problem: Show that there are infinitely many pairs $(m, n)$ of natural numbers $m, n \ge 2$, for $m^m- 1$  is divisible by $n$ and $n^n- 1$ is divisible by $m$.[

Proof: Aa easy and cute! I hope the construction is correct.

Let $n=m+1, m$ is even. So $$n \equiv 1 \mod m \implies n^n\equiv 1\mod m \implies n^n-1\equiv 0\mod m.$$

Since $m$ is even, $$n|m^2-1|m^m-1$$

For example, take any even $m,$ say $6,$ so $n=7.$

We have $$ 7|6^2-1|6^6-1,~~6|7^7-1=24|5.$$

Problem: Solve the equation $5^x7^y+4=3^z$ in nonnegative integers.

Proof:  No solution 

We have $$5^x7^y+4=3^z.$$

By $\mod 3,$ we get that $x$ is odd. 

By $\mod 4,$ we get that $y,z$ are of the same parity.

By $\mod 7,$ we get that $z$ is even and so is $y.$ Let $z=2k.$

Hence, $$5^x\cdot 7^y= (3^k-2)(3^k+2).$$

Since $$\gcd(3^k-2),(3^k+2)=1\implies 3^k-2=5^x \text{or} 3^k+2=5^x$$

Case 1:  $$ 3^k-2=5^x$$

Clearly $k>1,$ taking $\mod 9$ and using the fact that $x$ is odd, so $5^x\equiv 5,8,2 \mod 9.$

Not possible ( except $k=1,x=0$ which eventually doesn't work when we try it in $3^k+2=7^y.$ )

(another way is just $\mod 3$)

Case 2: $$3^k+2=5^x, 3^k-2=7^y$$

Clearly in $3^k+2=5^x,$ we get by $\mod 8 \implies k$ is odd.

But $\mod 7$ in $3^k-2=7^y$ gives $k$ is even.

Not possible.

Problem: Let $n$ be an integer. Show that a natural number $k$ can be found for which, the following applies with a suitable choice  of signs: $$n = \pm 1^2 \pm 2^2  \pm 3^2  \pm  ... \pm k^2$$

Proof: Pretty nice!

Note that if we represent $0$ and positive integer $n,$ then we are done as we can just change the signs.

$$ 0=1-4-(9-16)-(25-36)+(49-25)$$

$$ 1= 1$$

$$ 2= -1-4-9+16$$

$$ 3=-1+4$$

And then we use the fact that $$n^2-(n+1)^2-(n+2)^2+(n+3)^2=4.$$

P.S. My exam ended yesterday, didn't do that well in math and chem. So will focus on that! I did these problems during this 1 week.

P.S.S. From today onwards, I will officially start working on the Otis units! I don't think so I will be able to do a lot of units, but ye let's see. 

P.S.S.S. I think I have to update the pages. Also, I am thinking of making a page called "post log." Which will help the users to see different posts! I will probably add it in the upcoming weeks!

Well, that's it for this week. Also, 6k views special( Nope, no memes)  coming in the next week! Hope you enjoyed it. Happy problem-solving!

Sunaina 💜