Your first proof practice plan

Substitute y = 6 − x, then complete the square.

x² + y² = x² + (6 − x)² = 2x² − 12x + 36 = 2(x − 3)² + 18. The square is nonnegative, so the expression is at least 18. It equals 18 exactly when x = 3; then y = 3. This pair satisfies the constraint, so the bound is attained.

Check your reasoning: A bound is not a minimum until you show that an allowed pair attains it.

Add 4 after moving the linear terms to the left.

The equation is equivalent to (x − 2)(y − 2) = 4. Put u = x − 2 and v = y − 2. Since x,y are positive integers, u,v are integers at least −1. A negative pair with product 4 would need (−1,−4), (−2,−2) or (−4,−1), all outside these bounds. Thus u,v are positive. Their ordered factor pairs are (1,4), (2,2), (4,1). Hence (x,y) = (3,6), (4,4), (6,3). Substitution checks each pair.

Check your reasoning: Listing a few solutions is not enough; the factor-pair argument excludes every other case.

First show x is nonzero. Divide the equation by x.

Zero does not satisfy the equation, so division by x is allowed. It gives x + 1/x = 3. Squaring gives x² + 2 + 1/x² = 9, so x² + 1/x² = 7. Square again: x⁴ + 2 + 1/x⁴ = 49. Therefore the requested value is 47.

Check your reasoning: Explain why a denominator cannot be zero before dividing.

Factorise. Which factors must occur among three consecutive integers?

n³ − n = (n − 1)n(n + 1). Among three consecutive integers, one is a multiple of 3 and at least one is even. Thus the product is divisible by 2 and by 3. As 2 and 3 are coprime, it is divisible by 6. The argument also covers zero and negative integers.

Check your reasoning: Divisibility by a and b implies divisibility by ab only when their gcd is 1; here that condition holds.

A common divisor divides 2(n + 2) − (2n + 1).

Let d be the gcd. Since d divides both given numbers, it divides 2(n + 2) − (2n + 1) = 3. Therefore d is 1 or 3. Both numbers are divisible by 3 exactly when n ≡ 1 (mod 3): then n + 2 ≡ 0 and 2n + 1 ≡ 0. Hence d = 3 for n ≡ 1 (mod 3), and d = 1 otherwise.

Check your reasoning: Prove both directions of the condition for gcd 3, not only that the gcd divides 3.

Reduce modulo 3; then look for a smaller solution.

A square is 0 or 1 modulo 3. To have x² + y² ≡ 0, both squares must be 0; thus 3 divides x and y. Write x = 3a and y = 3b. Substitution gives 3(a² + b²) = z², so 3 also divides z. Write z = 3c; then a² + b² = 3c². If any positive solution existed, choose one with the smallest positive z. The new positive solution has c = z/3 < z, a contradiction. Hence none exists.

Check your reasoning: This is descent: merely showing all three numbers are divisible by 3 is not yet a contradiction. The smaller positive solution completes it.

First find AD/AB, not AD/DB.

Since AB = AD + DB, the ratio AD:AB is 2:3. Parallel lines give equal corresponding angles, so triangles ADE and ABC are similar. Their linear scale factor is 2/3; the ratio of areas is its square, 4/9. Thus the area ratio is 4:9.

Check your reasoning: The given ratio 2:1 compares two parts of AB. It is not the scale factor of the triangles.

Put the two legs on the coordinate axes.

Choose coordinates A = (0,0), B = (b,0), C = (0,c), where b,c > 0. The midpoint is M = (b/2,c/2). Therefore MA² = b²/4 + c²/4. Also MB² = (b/2 − b)² + c²/4 = b²/4 + c²/4, and MC² = b²/4 + (c/2 − c)² = b²/4 + c²/4. The three nonnegative lengths have equal squares, so they are equal.

Check your reasoning: Coordinates must represent an arbitrary right triangle; choosing equal legs would prove only a special case.

The condition implies cyclicity, not four right angles.

The student replaces a sum condition by two individual values. A convex cyclic quadrilateral can have angles 80°, 100°, 100°, 80° in order: the opposite pairs each sum to 180°, but no angle is right. For a concrete construction, take four points on a circle whose successive arcs measure 80°, 120°, 80°, 80°. The inscribed-angle theorem gives those angles (up to starting order). Thus the conclusion does not follow.

Check your reasoning: A counterexample must satisfy the hypothesis and violate the conclusion. Opposite-angle sums alone do not force a rectangle.

Write each number as a power of 2 times an odd number.

Every positive integer has a unique form 2ᵏm with m odd and k ≥ 0. For numbers 1–10 the odd part m is one of 1,3,5,7,9, giving five boxes. Six chosen numbers force two into the same box. Write them 2ʳm and 2ˢm with r < s. Then the second divided by the first is 2ˢ⁻ʳ, an integer. Hence the first divides the second.

Check your reasoning: The boxes are defined by odd part, not by whether the number itself is odd.

Add everyone’s handshake count. How often does one handshake occur in that sum?

Each handshake contributes 1 to each of two people’s counts, so the sum of all counts is twice the total number of handshakes and is even. The even individual counts contribute an even sum. Therefore the sum of the odd counts must also be even. A sum of k odd integers has the same parity as k, so k is even.

Check your reasoning: The total number of people need not be even; the conclusion concerns only those with odd counts.

Track the parity of the number of tails.

A move changes the number of tails by +2, 0 or −2: it flips two heads, one head and one tail, or two tails. Thus the parity of the number of tails never changes. Initially it is 0, an even number. A position with exactly one tail has odd parity, so it is unreachable.

Check your reasoning: Cover every type of legal move before claiming an invariant. A long unsuccessful search is not a proof of impossibility.