Closing the gap: quest to understand prime numbers; Dr Vicky Neale

You and your research or you and your studies for competitive math exams

You and your research ( You and your studies) : By Richard Hamming, AT and T, Bell Labs mathematician;

Method of undetermined coefficients for PreRMO, PRMO and IITJEE Foundation maths

Find out when the expression $x^{3}+px^{2}+qx+r$ is exactly divisible by $x^{2}+ax+b$

Solution 1:

Let $x^{3}+px^{2}+qx+r=(x^{2}+ax+b)(Ax+B)$ where A and B are to be determined in terms of p, q, r, a and b. We can assume so because we know from the fundamental theorem of algebra that the if the LHS has to be of degree three in x, the remaining factor in RHS has to be linear in x.

So, expanding out the RHS of above, we get:

$x^{3}+px^{2}+qx+r=Ax^{3}+aAx^{2}+bAx+Bx^{2}+Bax+bB$

$x^{3}+px^{3}+qx+r=Ax^{3}+(aA+B)x^{2}+x(bA+aB)+bB$

We are saying that the above is true for all values of x: hence, coefficients of like powers of x on LHS and RHS are same; we equate them and get a system of equations:

$A=1$

$p=aA+B$

$bA+aB=q$

$bB=r$

Hence, we get $p=a+\frac{r}{b}$ and $bp-ba=r$ or that $b(p-a)=r$

Also, $b+aB=q$ so that $q=b+\frac{ar}{b}$ which means $q-b=\frac{a}{b}r$

but $\frac{r}{b}=B=p-a$ and hence, $q-b=\frac{a}{b}(p-a)$

So, the required conditions are $b(p-a)=r$ and $q-b=\frac{a}{b}(p-a)$ .

2) Find the condition that $x^{2}+px+q$ may be a perfect square.

Solution 2:

Let $x^{2}+px+q=(Ax+B)^{2}$ where A and B are to be determined in terms of p and q; finally, we obtain the relationship required between p and q for the above requirement.

$x^{2}+px+q=A^{2}x^{2}+B^{2}+2ABx$ which is true for all real values of x;

Hence, $A^{2}=1$ so $A=1$ or $A=-1$

Also, $B^{2}=q$ and hence, $B=\sqrt{q}$ or $B=-\sqrt{q}$

Also, $2AB=p$ so that $2\sqrt{q}=p$ so $q=\frac{p^{2}}{4}$ , which is the required condition.

3) To prove that $x^{4}+px^{3}+qx^{2}+rx+s$ is a perfect square if $(q-\frac{p^{2}}{4})^{2}=4s$ and $r^{2}=p^{2}s$ .

Proof 3:

Let $x^{4}+px^{3}+qx^{2}+rx+s=(Ax^{2}+Bx+C)^{2}$

$x^{4}+px^{3}+qx^{2}+rx+s=A^{2}x^{4}+B^{2}x^{2}+C^{2}+2ABx^{3}+2BCx+2ACx^{2}$

$A^{2}=1$

$2AB=p$

$q=B^{2}+2AC$

$2BC=r$

$C^{2}=s$

$A=1$ or $A=-1$

$2AB=p \longrightarrow 2B=p \longrightarrow B=\frac{p}{2}$

$q=B^{2}+2AC=\frac{p^{2}}{4}+2\times \sqrt{s} \longrightarrow (q-\frac{p^{2}}{4})^{2}=4s$

$2 \times \frac{p}{2} \times \sqrt{s}=r \longrightarrow r^{2}=p^{2}s$

More later,

Nalin Pithwa.

PS: Note in the method of undetermined coefficients, we create an identity expression which is true for all real values of x.

Axiomatic Method : A little explanation

Posted by Nalin Pithwa

I) Take an English-into-English dictionary (any other language will also do). Start with any word and note down any word occurring in its definition, as given in the dictionary. Take this new word and note down any word appearing in it until a vicious circle results. Prove that a vicious circle is unavoidable no matter which word one starts with , (Caution: the vicious circle may not always involve the original word).

For example, in geometry the word “point” is undefined. For example, in set theory, when we write or say : $a \in A$ ; the element “a” ‘belongs to’ “set A” —- the word “belong to” is not defined.

So, in all branches of math or physics especially, there are such “atomic” or “undefined” terms that one starts with.

After such terms come the “axioms” — statements which are assumed to be true; that is, statements whose proof is not sought.

The following are the axioms based on which equations are solved in algebra:

If to equals we add equals, we get equals.
If from equals we take equals, the remainders are equal.
If equals are multiplied by equals, the products are equal.
If equals are divided by equals (not zero), the quotients are equal.

More later,

Nalin Pithwa.

Check your mathematical induction concepts

Posted by Nalin Pithwa

Discuss the following “proof” of the (false) theorem:

If n is any positive integer and S is a set containing exactly n real numbers, then all the numbers in S are equal:

PROOF BY INDUCTION:

Step 1:

If $n=1$ , the result is evident.

Step 2: By the induction hypothesis the result is true when $n=k$ ; we must prove that it is correct when $n=k+1$ . Let S be any set containing exactly $k+1$ real numbers and denote these real numbers by $a_{1}, a_{2}, a_{3}, \ldots, a_{k}, a_{k+1}$ . If we omit $a_{k+1}$ from this list, we obtain exactly k numbers $a_{1}, a_{2}, \ldots, a_{k}$ ; by induction hypothesis these numbers are all equal:

$a_{1}=a_{2}= \ldots = a_{k}$ .

If we omit $a_{1}$ from the list of numbers in S, we again obtain exactly k numbers $a_{2}, \ldots, a_{k}, a_{k+1}$ ; by the induction hypothesis these numbers are all equal:

$a_{2}=a_{3}=\ldots = a_{k}=a_{k+1}$ .

It follows easily that all $k+1$ numbers in S are equal.

*************************************************************************************

Comments, observations are welcome 🙂

Regards,

Nalin Pithwa

Miscellaneous Algebra: pRMO, IITJEE foundation maths 2019

Posted by Nalin Pithwa

For the following tutorial problems, it helps to know/remember/understand/apply the following identities (in addition to all other standard/famous identities you learn in high school maths):

$a^{3}+b^{3}+c^{3}-3abc=(a+b+c)(a^{2}+b^{2}+c^{2}-ab-bc-ca)$

By the way, I hope you also know how to derive the above.Let me mention two methods to derive the above :

Method I: Using polynomial division in three variable, divide the dividend $a^{3}+b^{3}+c^{3}-3abc$ by the divisor $a+b+c$ .

Method II: Assume that $P(X)$ is a polynomial with roots a, b and c. So, we know by the fundamental theorem of algebra that $P(X)=(X-a)(X-b)(X-c)$ . Now, we also know that a, b and c satisfy P(X). Now, proceed further and complete the proof.

Let us now work on the tutorial problems below:

1) If $2s=a+b+c$ , prove that $\frac{1}{s-a} + \frac{1}{s-b} + \frac{1}{s-c} = \frac{abc}{s(s-a)(s-b)(s-c)}$

2) If $x^{2}+a^{2}=2(xy+yz+zu-y^{2}-z^{2})$ , prove that $x=y=z=u$ .

Prove the following identities:

3) $b(x^{3}+a^{3})+ax(x^{2}-a^{2})+a^{3}(x+a)=(a+b)(x+a)(x^{2}-ax+a^{2})$

4) $(ax+by)^{2}+(ay-bx)^{2}+c^{2}x^{2}+c^{2}y^{2}=(x^{2}+y^{2})(a^{2}+b^{2}+c^{2})$

5) $(x+y)^{3}+ 3(x+y)^{2}z+3(x+y)z^{2}+z^{3}=(x+z)^{3}+3(x+z)^{2}y+3(x+z)y^{2}+y^{3}$

6) $(a+b+c)(ab+bc+ca)-abc=(a+b)(b+c)(c+a)$

7) $(a+b+c)^{2}-a(b+c-a)-b(a+c-b)-c(a+b-c)=2(a^{2}+b^{2}+c^{2})$

8) $(x-y)^{3}+(x+y)^{3}+3(x-y)^{2}(x+y)+3(x+y)^{2}(x-y)=8x^{3}$

9) $x^{2}(y-z)+y^{2}(z-x)+z^{2}(x-y)+(y-z)(z-x)(z-y)=0$

10) $a^{3}(b-c)+b^{3}(c-a)+c^{3}(a-b)=-(b-c)(c-a)(a-b)(a+b+c)$

11) Prove that $(b-c)^{3}+(c-a)^{3}+(a-b)^{3}=3(b-c)(c-a)(a-b)$

12) If3 $2s=a+b+c$ , prove that $(s-a)^{2}+(s-b)^{2}+(s-c)^{2}+s^{2}=a^{2}+b^{2}+c^{2}$

13) If $2s=a+b+c$ , prove that $(s-a)^{3}+(s-b)^{3}+(s-c)^{3}+3abc=s^{3}$

14) If $2s=a+b+c$ , prove that $16s(s-a)(s-b)(s-c)=2b^{2}c^{2}+2c^{2}a^{2}+2a^{2}b^{2}-a^{4}-b^{4}-c^{4}$

15) If $2s=a+b+c$ , then prove that $2(s-a)(s-b)(s-c)+a(s-b)(s-c)+b(s-c)(s-a)+c(s-a)(s-b)=abc$

16) If $a+b+c=0$ , then prove that $(2a-b)^{3}+(2b-c)^{3}+(2c-a)^{3}=3(2a-b)(2b-c)(2c-a)$

17) If $a+b+c=0$ , then prove that $\frac{a^{2}}{2a^{2}+bc} + \frac{b^{2}}{2b^{2}+ca} + \frac{c^{2}}{2c^{2}+ab} =1$

18) Prove that $(x+y+z)^{3}+(x+y-z)^{3}+(x-y+z)^{3}+(x-y-z)^{3}=4x(x^{2}+3y^{2}+3z^{2})$

19) If $a+b+c=0$ prove that $(s+3a)^{3}-(s-3b)^{3}-(s-3c)^{3}-3(s-3a)(s-3b)(s-3c)=0$

20) If $X=b+c-2a$ , $Y=c+a-2b$ , $Z=a+b-2c$ , find the value of $X^{2}+Y^{2}+Z^{2}-3XYZ$

21) Prove that $(a-b)^{2}+(b-c)^{2}+(c-a)^{2}=2(c-b)(c-a)+2(b-a)(b-c)+2(a-b)(a-c)$

22) Prove that $a^{2}(b^{3}-c^{3})+b^{2}(c^{3}-a^{3})+c^{2}(a^{3}-b^{3})=(a-b)(b-c)(c-a)(ab+bc+ca)=a^{2}(b-c)^{3}+b^{2}(c-a)^{3}+c^{2}(a-b)^{3} = -[a^{2}b^{2}(a-b)+b^{2}c^{2}(b-c)+c^{2}a^{2}(c-a)]$

23) if $(a+b)^{2}+(b+c)^{2}+(c+a)^{2}=4(ab+bc+cd)$ , prove that $a=b=c=d$ .

24) If $x=a+d$ , $y=b+d$ , $z=c+d$ , prove that $x^{2}+y^{2}+z^{2}-yz-zx-xy=a^{2}+b^{2}+c^{2}-bc-ca-ab$

25) If $a+b+c=3$ , prove that $\frac{1}{b^{2}+c^{2}-a^{2}}+ \frac{1}{c^{2}+a^{2}-b^{2}} + \frac{1}{a^{2}+b^{2}-c^{2}}=0$

26) If $a+b+c=0$ , simplify: $\frac{b+c}{bc}(b^{2}+c^{2}-a^{2}) + \frac{c+a}{ca} (c^{2}+a^{2}-b^{2})+ \frac{a+b}{ab}(a^{2}+b^{2}-c^{2})$

27) Prove that the equation $(x-a)^{2}+(y-b)^{2}+(a^{2}+b^{2}-1)(x^{2}+y^{2}-1)=0$ is equivalent to the equation $(ax+by-1)^{2}+(bx-ay)^{2}=0$ , hence show that the only possible values of x and y are: $\frac{a}{a^{2}+b^{2}}$ , $\frac{b}{a^{2}+b^{2}}$

28) If $2(x^{2}+a^{2}-ax)(y^{2}+b^{2}-by)=x^{2}y^{2}+a^{2}b^{2}$ , prove that $(x-a)^{2}(y-b)^{2}+(bx-ay)^{2}=0$ and therefore that $a=x$ and $y=b$ are the only possible solutions.

Good luck for the PreRMo August 2019 !!

Regards,

Nalin Pithwa

Cyclic expressions, fractions: Pre RMO, PRMO, IITJEE foundation 2019

Posted by Nalin Pithwa

In order to solve the following tutorial sheet, it helps to solve/understand and then apply the following beautiful cyclic relations or identities:

(Note if these look new to you, then you need to check the truth of all them; if all are v v familiar to you, just go ahead and crack the tutorial sheet below):

Core Identities in Cyclic Expressions:
1) $(b-c)+(c-a)+(a-b)=0$
2) $a(b-c)+b(c-a)+c(a-b)=0$
3) $a^{2}(b-c)+b^{2}(c-a)+c^{2}(a-b)=-(a-b)(b-c)(c-a)$
4) $bc(b-c)+ca(c-a)+ab(a-b)=-(a-b)(b-c)(c-a)$
5) $a(b^{2}-c^{2})+b(c^{2}-a^{2})+c(a^{2}-b^{2})=(a-b)(b-c)(c-a)$

Solve or simplify the following:

1) $\frac{a}{(a-b)(a-c)} + \frac{b}{(b-c)(b-a)} + \frac{c}{(c-a)(c-b)}$
2) $\frac{bc}{(a-b)(a-c)} + \frac{ca}{(b-c)(b-a)} + \frac{ab}{(c-a)(c-b)}$
3) $\frac{a^{2}}{(a-b)(a-c)} + \frac{b^{2}}{(b-c)(b-a)} + \frac{c^{2}}{(c-a)(c-b)}$
4) $\frac{a^{3}}{(a-b)(a-c)} + \frac{b^{3}}{(b-c)(b-a)} + \frac{c^{3}}{(c-a)(c-b)}$
5) $\frac{a(b+c)}{(a-b)(c-a)} + \frac{b(a+c)}{(a-b)(b-c)} + \frac{a(a+b)}{(c-a)(b-c)}$
6) $\frac{1}{a(a-b)(a-c)} + \frac{1}{b(b-c)(b-a)} + \frac{1}{c(c-a)(c-b)}$
7) $\frac{bc}{a(a^{2}-b^{2})(a^{2}-c^{2})} + \frac{ca}{b(b^{2}-c^{2})(b^{2}-a^{2})} + \frac{ab}{c(c^{2}-a^{2})(c^{2}-b^{2})}$
8) $\frac{(x-b)(x-c)}{(a-b)(a-c)} + \frac{(x-c)(x-a)}{(b-c)(b-a)} + \frac{(x-a)(x-b)}{(c-a)(c-b)}$
9) $\frac{bc(a+d)}{(a-b)(a-c)} + \frac{ca(b+d)}{(b-c)(b-a)} + \frac{ab(c+d)}{(c-a)(c-b)}$
10) $\frac{1}{(a-b)(a-c)(x-a)} + \frac{1}{(b-c)(b-a)(x-b)} + \frac{1}{(c-a)(c-b)(x-c)}$
11) $\frac{a^{2}}{(a-b)(a-c)(x+a)} + \frac{b^{2}}{(b-c)(b-a)(x+b)} + \frac{c^{2}}{(c-a)(c-b)(x+c)}$
12) $a^{2}\frac{(a+b)(a+c)}{(a-b)(a-c)} + b^{2}\frac{(b+c)(b+a)}{(b-c)(b-a)} + c^{2}\frac{(c+a)(c+b)}{(c-a)(c-b)}$
13) $\frac{a^{3}(b-c)+b^{3}(c-a)+c^{3}(a-b)}{(b-c)^{3}+(c-a)^{3}+(a-b)^{3}}$
14) $\frac{a^{2}(b-c)+b^{2}(c-a)+c^{2}(a-b)+2(c-a)(a-b)(b-c)}{(b-c)^{3}+(c-a)^{3}+(a-b)^{3}}$
15) $\frac{a^{3}(b-c)+b^{3}(c-a)+c^{3}(a-b)}{a^{2}(b-c)+b^{2}(c-a)+c^{2}(a-b)}$
16) $\frac{a^{2}(b-c)^{3}+b^{2}(c-a)^{3}+c^{2}(a-b)^{3}}{(a-b)(b-c)(c-a)}$
17) $\frac{\frac{b-c}{a} + \frac{c-a}{b} + \frac{a-b}{c}}{\frac{1}{a}(\frac{1}{b^{2}}-\frac{1}{c^{2}})+\frac{1}{b}(\frac{1}{c^{2}}-\frac{1}{a^{2}})+\frac{1}{c}(\frac{1}{a^{2}}-\frac{1}{b^{2}})}$ ^
18) $\frac{a^{2}(\frac{1}{a^{2}}-\frac{1}{b^{2}})+b^{2}(\frac{1}{a^{2}}-\frac{1}{c^{2}})+c^{2}(\frac{1}{b^{2}}-\frac{1}{a^{2}})}{\frac{1}{bc}(\frac{1}{c}-\frac{1}{b})+\frac{1}{ca}(\frac{1}{a}-\frac{1}{c})+\frac{1}{ab}(\frac{1}{b}-\frac{1}{c})}$
19) $\frac{a}{(a-b)(a-c)(x-a)} + \frac{b}{(b-c)(b-a)(x-b)} + \frac{c}{(c-a)(c-b)(x-c)}$

More later,
Nalin Pithwa